Class 10 Mathematics Chapter 8 Question Answer | Introduction to Trigonometry | English Medium

Class 10 Mathematics Chapter 8 Question Answer | Introduction to Trigonometry | English Medium | ASSEB

Welcome to HSLC Guru

Hello dear student! Welcome to HSLC Guru. In this lesson you will find the complete English-medium solutions for ASSEB (Assam State School Education Board) Class 10 Mathematics, Chapter 8 — Introduction to Trigonometry. Trigonometry is one of the oldest branches of mathematics, originally invented by astronomers to study the relationship between the sides and angles of a right-angled triangle. The word trigonometry itself comes from three Greek roots — tri (three), gon (sides) and metron (measure) — literally meaning “the measurement of a three-sided figure.” This chapter introduces the six trigonometric ratios, the values of these ratios for the standard angles $0°, 30°, 45°, 60°$ and $90°$, the relationship between trigonometric ratios of complementary angles, and the three fundamental trigonometric identities. The full textbook exercises 8.1, 8.2, 8.3 and 8.4 are solved here step-by-step, followed by a section of additional questions that frequently appear in HSLC examinations.

Summary

In a right-angled triangle, with reference to one of the acute angles $\theta$, the side opposite to $\theta$ is called the perpendicular (P) or opposite side, the side adjacent to $\theta$ (other than the hypotenuse) is called the base (B) or adjacent side, and the longest side opposite the right angle is the hypotenuse (H). The six trigonometric ratios — sine, cosine, tangent, cosecant, secant and cotangent — are defined as the ratios between these three sides. These ratios are pure numbers (with no units) and depend only on the angle $\theta$, not on the size of the triangle. The ratios obey three fundamental identities — $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$ — that hold for every value of $\theta$ for which the expressions are defined. Trigonometric ratios of complementary angles satisfy the rule that “the sine of an angle equals the cosine of its complement,” and similarly for the other co-functions. The standard-angle table memorised in this chapter — values at $0°, 30°, 45°, 60°, 90°$ — is the toolbox used throughout the rest of the syllabus. Mastery of this chapter is the foundation for Chapter 9 (Some Applications of Trigonometry, including heights and distances).

Right-Angled Triangle: Sides Relative to an Angle

Consider a right-angled triangle $ABC$ with the right angle at $B$, and the angle $\theta$ at vertex $A$. With reference to angle $\theta = \angle A$:

$BC$ is the side opposite to $\theta$ — call it Perpendicular (P).
$AB$ is the side adjacent to $\theta$ (not the hypotenuse) — call it Base (B).
$AC$ is the side opposite the right angle — the Hypotenuse (H).

By the Pythagoras Theorem,

$$H^2 = P^2 + B^2 \quad\text{i.e.}\quad AC^2 = BC^2 + AB^2$$

The Six Trigonometric Ratios

For the acute angle $\theta$ in the right triangle above, the six trigonometric ratios are defined as follows.

Ratio	Definition	In terms of P, B, H
$\sin\theta$	Perpendicular / Hypotenuse	$\dfrac{P}{H}$
$\cos\theta$	Base / Hypotenuse	$\dfrac{B}{H}$
$\tan\theta$	Perpendicular / Base	$\dfrac{P}{B}$
$\csc\theta$	Hypotenuse / Perpendicular	$\dfrac{H}{P}$
$\sec\theta$	Hypotenuse / Base	$\dfrac{H}{B}$
$\cot\theta$	Base / Perpendicular	$\dfrac{B}{P}$

Reciprocal relations:

$$\csc\theta = \frac{1}{\sin\theta}, \qquad \sec\theta = \frac{1}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta}$$

Quotient relations:

$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

Standard-Angle Table

The values of the six trigonometric ratios at the angles $0°, 30°, 45°, 60°$ and $90°$ are used everywhere in trigonometry and must be memorised.

$\theta$	$0°$	$30°$	$45°$	$60°$	$90°$
$\sin\theta$	$0$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt2}$	$\dfrac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{\sqrt2}$	$\dfrac{1}{2}$	$0$
$\tan\theta$	$0$	$\dfrac{1}{\sqrt3}$	$1$	$\sqrt3$	not defined
$\csc\theta$	not defined	$2$	$\sqrt2$	$\dfrac{2}{\sqrt3}$	$1$
$\sec\theta$	$1$	$\dfrac{2}{\sqrt3}$	$\sqrt2$	$2$	not defined
$\cot\theta$	not defined	$\sqrt3$	$1$	$\dfrac{1}{\sqrt3}$	$0$

Memory tip: Write the sine row as $\sqrt{0}/2,\ \sqrt{1}/2,\ \sqrt{2}/2,\ \sqrt{3}/2,\ \sqrt{4}/2$. The cosine row is the same numbers in reverse. Tangent = Sine / Cosine.

Trigonometric Ratios of Complementary Angles

Two angles are complementary if they add up to $90°$. In a right-angled triangle, the two acute angles are always complementary. If $\theta$ is one acute angle, then the other is $(90°-\theta)$. Their ratios are linked by the following rules.

$$\sin(90°-\theta) = \cos\theta \qquad \cos(90°-\theta) = \sin\theta$$
$$\tan(90°-\theta) = \cot\theta \qquad \cot(90°-\theta) = \tan\theta$$
$$\sec(90°-\theta) = \csc\theta \qquad \csc(90°-\theta) = \sec\theta$$

In words — sine becomes cosine, tangent becomes cotangent, and secant becomes cosecant when the angle is replaced by its complement (and vice-versa).

Trigonometric Identities

The three fundamental identities, valid for every angle $\theta$ in the range where the ratios are defined:

Identity	Equivalent forms
$\sin^2\theta + \cos^2\theta = 1$	$\sin^2\theta = 1 – \cos^2\theta$ ; $\cos^2\theta = 1 – \sin^2\theta$
$1 + \tan^2\theta = \sec^2\theta$	$\sec^2\theta – \tan^2\theta = 1$ ; $\tan^2\theta = \sec^2\theta – 1$
$1 + \cot^2\theta = \csc^2\theta$	$\csc^2\theta – \cot^2\theta = 1$ ; $\cot^2\theta = \csc^2\theta – 1$

The first identity follows directly from Pythagoras’ theorem applied to a right triangle with hypotenuse $1$ and acute angle $\theta$. The other two are obtained by dividing the first identity throughout by $\cos^2\theta$ and $\sin^2\theta$ respectively.

Exercise 8.1

Question 1: In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine: (i) $\sin A, \cos A$ ; (ii) $\sin C, \cos C$.

Answer: By Pythagoras’ theorem,

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 7^2} = \sqrt{576+49} = \sqrt{625} = 25 \text{ cm}$$

(i) For angle $A$: opposite side $= BC = 7$, hypotenuse $= AC = 25$, adjacent $= AB = 24$.

$\sin A = \dfrac{BC}{AC} = \dfrac{7}{25}$ ; $\cos A = \dfrac{AB}{AC} = \dfrac{24}{25}$.

(ii) For angle $C$: opposite side $= AB = 24$, adjacent $= BC = 7$, hypotenuse $= AC = 25$.

$\sin C = \dfrac{AB}{AC} = \dfrac{24}{25}$ ; $\cos C = \dfrac{BC}{AC} = \dfrac{7}{25}$.

Question 2: In the figure, find $\tan P – \cot R$. ($PR = 13$ cm, $PQ = 12$ cm, right angle at $Q$.)

Answer: $QR = \sqrt{PR^2 – PQ^2} = \sqrt{169-144} = \sqrt{25} = 5$ cm.

$\tan P = \dfrac{QR}{PQ} = \dfrac{5}{12}$ and $\cot R = \dfrac{QR}{PQ} = \dfrac{5}{12}$.

$\therefore\ \tan P – \cot R = \dfrac{5}{12} – \dfrac{5}{12} = 0$.

Question 3: If $\sin A = \dfrac{3}{4}$, calculate $\cos A$ and $\tan A$.

Answer: Let in $\triangle ABC$, right-angled at $B$, $\angle A$ be the acute angle.

$\sin A = \dfrac{BC}{AC} = \dfrac{3}{4}$. Take $BC = 3k, AC = 4k$ for some constant $k>0$.

By Pythagoras, $AB = \sqrt{AC^2 – BC^2} = \sqrt{16k^2 – 9k^2} = k\sqrt{7}$.

$\cos A = \dfrac{AB}{AC} = \dfrac{k\sqrt7}{4k} = \dfrac{\sqrt7}{4}$ ; $\tan A = \dfrac{BC}{AB} = \dfrac{3k}{k\sqrt7} = \dfrac{3}{\sqrt7}$.

Question 4: Given $15\cot A = 8$, find $\sin A$ and $\sec A$.

Answer: $\cot A = \dfrac{8}{15}$, so in $\triangle ABC$ right-angled at $B$, with $\angle A$ acute, $\dfrac{AB}{BC} = \dfrac{8}{15}$. Let $AB = 8k, BC = 15k$.

$AC = \sqrt{AB^2 + BC^2} = \sqrt{64k^2+225k^2} = \sqrt{289k^2} = 17k$.

$\sin A = \dfrac{BC}{AC} = \dfrac{15}{17}$ ; $\sec A = \dfrac{AC}{AB} = \dfrac{17}{8}$.

Question 5: Given $\sec\theta = \dfrac{13}{12}$, calculate all other trigonometric ratios.

Answer: $\sec\theta = \dfrac{H}{B} = \dfrac{13}{12}$. So $H = 13k, B = 12k$ and $P = \sqrt{H^2 – B^2} = \sqrt{169k^2 – 144k^2} = 5k$.

$\sin\theta = \dfrac{P}{H} = \dfrac{5}{13}$ ; $\cos\theta = \dfrac{B}{H} = \dfrac{12}{13}$ ; $\tan\theta = \dfrac{P}{B} = \dfrac{5}{12}$ ; $\csc\theta = \dfrac{13}{5}$ ; $\cot\theta = \dfrac{12}{5}$.

Question 6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.

Answer: Let $\triangle ABC$ have $\cos A = \dfrac{AC}{AB}$ and $\cos B = \dfrac{BC}{AB}$ in some configuration; more directly, drop the perpendicular from $C$ to $AB$ at $D$. Then $\cos A = \dfrac{AD}{AC}$ and $\cos B = \dfrac{BD}{BC}$.

Given $\cos A = \cos B$, i.e. $\dfrac{AD}{AC} = \dfrac{BD}{BC}$. By comparing right triangles $ACD$ and $BCD$ with the common altitude $CD$, the only way the cosines (adjacent/hypotenuse) of two acute angles can be equal is if the angles themselves are equal. Hence $\angle A = \angle B$.

Question 7: If $\cot\theta = \dfrac{7}{8}$, evaluate (i) $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ ; (ii) $\cot^2\theta$.

Answer: $\cot\theta = \dfrac{B}{P} = \dfrac{7}{8}$. So $B = 7k, P = 8k, H = \sqrt{49k^2+64k^2} = k\sqrt{113}$.

(i) Numerator $= 1 – \sin^2\theta = \cos^2\theta$. Denominator $= 1 – \cos^2\theta = \sin^2\theta$. So expression $= \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \left(\dfrac{7}{8}\right)^2 = \dfrac{49}{64}$.

(ii) $\cot^2\theta = \dfrac{49}{64}$.

Question 8: If $3\cot A = 4$, check whether $\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A – \sin^2 A$ or not.

Answer: $\cot A = \dfrac{4}{3} \Rightarrow \tan A = \dfrac{3}{4}$. Take base $= 4k$, perpendicular $= 3k$, hypotenuse $= 5k$.

$\sin A = \dfrac{3}{5}, \cos A = \dfrac{4}{5}, \tan A = \dfrac{3}{4}$.

LHS $= \dfrac{1 – \tan^2 A}{1 + \tan^2 A} = \dfrac{1 – 9/16}{1 + 9/16} = \dfrac{7/16}{25/16} = \dfrac{7}{25}$.

RHS $= \cos^2 A – \sin^2 A = \dfrac{16}{25} – \dfrac{9}{25} = \dfrac{7}{25}$.

$\therefore$ LHS $=$ RHS. Verified.

Question 9: In $\triangle ABC$ right-angled at $B$, if $\tan A = \dfrac{1}{\sqrt3}$, find the value of (i) $\sin A\cos C + \cos A\sin C$ ; (ii) $\cos A\cos C – \sin A\sin C$.

Answer: $\tan A = \dfrac{1}{\sqrt3} \Rightarrow A = 30°$, hence $C = 60°$.

$\sin A = \dfrac{1}{2}, \cos A = \dfrac{\sqrt3}{2}, \sin C = \dfrac{\sqrt3}{2}, \cos C = \dfrac{1}{2}$.

(i) $\sin A\cos C + \cos A\sin C = \dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} = \dfrac{1}{4} + \dfrac{3}{4} = 1$.

(ii) $\cos A\cos C – \sin A\sin C = \dfrac{\sqrt3}{2}\cdot\dfrac{1}{2} – \dfrac{1}{2}\cdot\dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{4} – \dfrac{\sqrt3}{4} = 0$.

Question 10: In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.

Answer: Let $QR = x$, then $PR = 25 – x$. By Pythagoras, $PR^2 = PQ^2 + QR^2$:

$(25-x)^2 = 25 + x^2 \Rightarrow 625 – 50x + x^2 = 25 + x^2 \Rightarrow 50x = 600 \Rightarrow x = 12$ cm.

So $QR = 12$ cm and $PR = 13$ cm.

$\sin P = \dfrac{QR}{PR} = \dfrac{12}{13}$ ; $\cos P = \dfrac{PQ}{PR} = \dfrac{5}{13}$ ; $\tan P = \dfrac{QR}{PQ} = \dfrac{12}{5}$.

Question 11: State whether the following are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than $1$.
(ii) $\sec A = \dfrac{12}{5}$ for some value of angle $A$.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
(iv) $\cot A$ is the product of $\cot$ and $A$.
(v) $\sin\theta = \dfrac{4}{3}$ for some angle $\theta$.

Answer:

(i) False. $\tan 60° = \sqrt3 > 1$ and $\tan A$ can be any non-negative real value as $A$ ranges over $[0°, 90°)$.

(ii) True. $\sec A \geq 1$ for $0° \leq A < 90°$. Since $\dfrac{12}{5} = 2.4 > 1$, there exists an angle $A$ with $\sec A = \dfrac{12}{5}$.

(iii) False. $\cos A$ is the abbreviation for cosine of angle $A$, not cosecant. Cosecant is written as $\csc A$ (or $\operatorname{cosec} A$).

(iv) False. $\cot A$ is a single symbol — the cotangent of $A$. It is not a multiplication “cot $\times$ A.”

(v) False. Since $-1 \leq \sin\theta \leq 1$ for all angles $\theta$, $\sin\theta = \dfrac{4}{3} > 1$ is impossible.

Exercise 8.2

Question 1: Evaluate the following.
(i) $\sin 60°\cos 30° + \sin 30°\cos 60°$
(ii) $2\tan^2 45° + \cos^2 30° – \sin^2 60°$
(iii) $\dfrac{\cos 45°}{\sec 30° + \csc 30°}$
(iv) $\dfrac{\sin 30° + \tan 45° – \csc 60°}{\sec 30° + \cos 60° + \cot 45°}$
(v) $\dfrac{5\cos^2 60° + 4\sec^2 30° – \tan^2 45°}{\sin^2 30° + \cos^2 30°}$

Answer:

(i) $= \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{3}{4} + \dfrac{1}{4} = 1$.

(ii) $= 2(1)^2 + \left(\dfrac{\sqrt3}{2}\right)^2 – \left(\dfrac{\sqrt3}{2}\right)^2 = 2 + \dfrac{3}{4} – \dfrac{3}{4} = 2$.

(iii) $\sec 30° = \dfrac{2}{\sqrt3}$ and $\csc 30° = 2$. Numerator $= \cos 45° = \dfrac{1}{\sqrt2}$.

$$\frac{1/\sqrt2}{2/\sqrt3 + 2} = \frac{1/\sqrt2}{\frac{2 + 2\sqrt3}{\sqrt3}} = \frac{\sqrt3}{\sqrt2\,(2+2\sqrt3)} = \frac{\sqrt3}{2\sqrt2(1+\sqrt3)}$$

Rationalising by multiplying numerator and denominator by $(\sqrt3 – 1)$ after re-writing: more cleanly, multiply numerator and denominator by $\sqrt3$ to clear: result $= \dfrac{3\sqrt2}{8} \cdot \dfrac{\sqrt3-1}{\sqrt3-1}$ — performing the standard simplification gives the final form $\dfrac{3\sqrt2 – \sqrt6}{8}$. Multiplying numerator and denominator of the original fraction by $\sqrt3$:

$$=\frac{\sqrt3}{\sqrt2(2+2\sqrt3)} \cdot \frac{\sqrt3-1}{\sqrt3-1} \cdot \frac{1}{1} = \frac{3\sqrt2 – \sqrt6}{8}$$

(iv) $\sin 30° = \dfrac{1}{2},\ \tan 45° = 1,\ \csc 60° = \dfrac{2}{\sqrt3},\ \sec 30° = \dfrac{2}{\sqrt3},\ \cos 60° = \dfrac{1}{2},\ \cot 45° = 1$.

Numerator $= \dfrac{1}{2} + 1 – \dfrac{2}{\sqrt3} = \dfrac{3}{2} – \dfrac{2}{\sqrt3} = \dfrac{3\sqrt3 – 4}{2\sqrt3}$.

Denominator $= \dfrac{2}{\sqrt3} + \dfrac{1}{2} + 1 = \dfrac{2}{\sqrt3} + \dfrac{3}{2} = \dfrac{4 + 3\sqrt3}{2\sqrt3}$.

$\therefore$ Ratio $= \dfrac{3\sqrt3 – 4}{4 + 3\sqrt3}$. Rationalising by $(3\sqrt3 – 4)$:

$$= \frac{(3\sqrt3 – 4)(3\sqrt3 – 4)}{(4 + 3\sqrt3)(3\sqrt3 – 4)} = \frac{(3\sqrt3-4)^2}{27 – 16} = \frac{27 – 24\sqrt3 + 16}{11} = \frac{43 – 24\sqrt3}{11}$$

(v) Denominator $= \sin^2 30° + \cos^2 30° = 1$.

Numerator $= 5\cdot\dfrac{1}{4} + 4\cdot\dfrac{4}{3} – 1 = \dfrac{5}{4} + \dfrac{16}{3} – 1 = \dfrac{15 + 64 – 12}{12} = \dfrac{67}{12}$.

$\therefore$ Required value $= \dfrac{67}{12}$.

Question 2: Choose the correct option and justify your choice.
(i) $\dfrac{2\tan 30°}{1 + \tan^2 30°} = $ (A) $\sin 60°$ (B) $\cos 60°$ (C) $\tan 60°$ (D) $\sin 30°$
(ii) $\dfrac{1 – \tan^2 45°}{1 + \tan^2 45°} = $ (A) $\tan 90°$ (B) $1$ (C) $\sin 45°$ (D) $0$
(iii) $\sin 2A = 2\sin A$ is true when $A=$ (A) $0°$ (B) $30°$ (C) $45°$ (D) $60°$
(iv) $\dfrac{2\tan 30°}{1 – \tan^2 30°} = $ (A) $\cos 60°$ (B) $\sin 60°$ (C) $\tan 60°$ (D) $\sin 30°$

Answer:

(i) $\tan 30° = \dfrac{1}{\sqrt3}$. $\dfrac{2/\sqrt3}{1 + 1/3} = \dfrac{2/\sqrt3}{4/3} = \dfrac{2}{\sqrt3}\cdot\dfrac{3}{4} = \dfrac{3}{2\sqrt3} = \dfrac{\sqrt3}{2} = \sin 60°$. (A)

(ii) $\tan 45° = 1$. $\dfrac{1 – 1}{1 + 1} = 0$. (D)

(iii) At $A = 0°$: $\sin 0° = 0 = 2\sin 0°$. Verified. (A)

(iv) $\dfrac{2/\sqrt3}{1 – 1/3} = \dfrac{2/\sqrt3}{2/3} = \dfrac{2}{\sqrt3}\cdot\dfrac{3}{2} = \dfrac{3}{\sqrt3} = \sqrt3 = \tan 60°$. (C)

Question 3: If $\tan(A+B) = \sqrt3$ and $\tan(A-B) = \dfrac{1}{\sqrt3}, 0° < A+B \leq 90°, A > B$, find $A$ and $B$.

Answer: $\tan(A+B) = \sqrt3 \Rightarrow A+B = 60°$. $\tan(A-B) = \dfrac{1}{\sqrt3} \Rightarrow A-B = 30°$.

Adding: $2A = 90° \Rightarrow A = 45°$. Subtracting: $2B = 30° \Rightarrow B = 15°$.

Question 4: State whether the following are true or false. Justify.
(i) $\sin(A + B) = \sin A + \sin B$.
(ii) The value of $\sin\theta$ increases as $\theta$ increases.
(iii) The value of $\cos\theta$ increases as $\theta$ increases.
(iv) $\sin\theta = \cos\theta$ for all values of $\theta$.
(v) $\cot A$ is not defined for $A = 0°$.

Answer:

(i) False. Take $A = B = 30°$: $\sin 60° = \dfrac{\sqrt3}{2}$ but $\sin 30° + \sin 30° = 1$. They are not equal.

(ii) True. As $\theta$ increases from $0°$ to $90°$, $\sin\theta$ increases from $0$ to $1$.

(iii) False. As $\theta$ increases from $0°$ to $90°$, $\cos\theta$ decreases from $1$ to $0$.

(iv) False. $\sin\theta = \cos\theta$ only at $\theta = 45°$, not at all angles.

(v) True. $\cot A = \dfrac{\cos A}{\sin A}$ and $\sin 0° = 0$, so $\cot 0°$ is undefined (division by zero).

Exercise 8.3

Question 1: Evaluate.
(i) $\dfrac{\sin 18°}{\cos 72°}$
(ii) $\dfrac{\tan 26°}{\cot 64°}$
(iii) $\cos 48° – \sin 42°$
(iv) $\csc 31° – \sec 59°$

Answer:

(i) $\cos 72° = \cos(90°-18°) = \sin 18°$. $\therefore \dfrac{\sin 18°}{\sin 18°} = 1$.

(ii) $\cot 64° = \cot(90°-26°) = \tan 26°$. $\therefore \dfrac{\tan 26°}{\tan 26°} = 1$.

(iii) $\sin 42° = \sin(90°-48°) = \cos 48°$. $\therefore \cos 48° – \cos 48° = 0$.

(iv) $\sec 59° = \sec(90°-31°) = \csc 31°$. $\therefore \csc 31° – \csc 31° = 0$.

Question 2: Show that.
(i) $\tan 48°\tan 23°\tan 42°\tan 67° = 1$
(ii) $\cos 38°\cos 52° – \sin 38°\sin 52° = 0$

Answer:

(i) $\tan 67° = \tan(90°-23°) = \cot 23° = \dfrac{1}{\tan 23°}$. Similarly $\tan 42° = \cot 48° = \dfrac{1}{\tan 48°}$.

$\therefore$ LHS $= \tan 48°\cdot\tan 23°\cdot\dfrac{1}{\tan 48°}\cdot\dfrac{1}{\tan 23°} = 1$.

(ii) $\cos 52° = \cos(90°-38°) = \sin 38°$ and $\sin 52° = \sin(90°-38°) = \cos 38°$.

$\therefore$ LHS $= \cos 38°\sin 38° – \sin 38°\cos 38° = 0$. Hence proved.

Question 3: If $\tan 2A = \cot(A – 18°)$, where $2A$ is an acute angle, find the value of $A$.

Answer: $\cot(A – 18°) = \tan(90° – (A – 18°)) = \tan(108° – A)$.

$\therefore \tan 2A = \tan(108° – A) \Rightarrow 2A = 108° – A \Rightarrow 3A = 108° \Rightarrow A = 36°$.

Question 4: If $\tan A = \cot B$, prove that $A + B = 90°$.

Answer: $\cot B = \tan(90° – B)$. So $\tan A = \tan(90° – B) \Rightarrow A = 90° – B \Rightarrow A + B = 90°$. Proved.

Question 5: If $\sec 4A = \csc(A – 20°)$, where $4A$ is an acute angle, find $A$.

Answer: $\csc(A – 20°) = \sec(90° – (A – 20°)) = \sec(110° – A)$.

$\therefore \sec 4A = \sec(110° – A) \Rightarrow 4A = 110° – A \Rightarrow 5A = 110° \Rightarrow A = 22°$.

Question 6: If $A, B, C$ are interior angles of $\triangle ABC$, then show that $\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2}$.

Answer: In $\triangle ABC$, $A + B + C = 180°$, so $B + C = 180° – A$, i.e. $\dfrac{B+C}{2} = 90° – \dfrac{A}{2}$.

$\therefore \sin\dfrac{B+C}{2} = \sin\left(90° – \dfrac{A}{2}\right) = \cos\dfrac{A}{2}$. Proved.

Question 7: Express $\sin 67° + \cos 75°$ in terms of trigonometric ratios of angles between $0°$ and $45°$.

Answer: $\sin 67° = \sin(90° – 23°) = \cos 23°$ ; $\cos 75° = \cos(90° – 15°) = \sin 15°$.

$\therefore \sin 67° + \cos 75° = \cos 23° + \sin 15°$.

Exercise 8.4

Question 1: Express the trigonometric ratios $\sin A, \sec A$ and $\tan A$ in terms of $\cot A$.

Answer: Use the identity $1 + \cot^2 A = \csc^2 A$.

$\csc A = \sqrt{1 + \cot^2 A}$, so $\sin A = \dfrac{1}{\csc A} = \dfrac{1}{\sqrt{1 + \cot^2 A}}$.

$\tan A = \dfrac{1}{\cot A}$.

For $\sec A$: $\sec^2 A = 1 + \tan^2 A = 1 + \dfrac{1}{\cot^2 A} = \dfrac{\cot^2 A + 1}{\cot^2 A}$.

$\therefore \sec A = \dfrac{\sqrt{1 + \cot^2 A}}{\cot A}$ (taking positive root for acute angle).

Question 2: Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Answer:

$\cos A = \dfrac{1}{\sec A}$.

$\sin^2 A = 1 – \cos^2 A = 1 – \dfrac{1}{\sec^2 A} = \dfrac{\sec^2 A – 1}{\sec^2 A}\Rightarrow \sin A = \dfrac{\sqrt{\sec^2 A – 1}}{\sec A}$.

$\tan A = \dfrac{\sin A}{\cos A} = \sqrt{\sec^2 A – 1}$.

$\csc A = \dfrac{1}{\sin A} = \dfrac{\sec A}{\sqrt{\sec^2 A – 1}}$.

$\cot A = \dfrac{1}{\tan A} = \dfrac{1}{\sqrt{\sec^2 A – 1}}$.

Question 3: Evaluate.
(i) $\dfrac{\sin^2 63° + \sin^2 27°}{\cos^2 17° + \cos^2 73°}$
(ii) $\sin 25°\cos 65° + \cos 25°\sin 65°$

Answer:

(i) $\sin 63° = \cos 27°$ so $\sin^2 63° + \sin^2 27° = \cos^2 27° + \sin^2 27° = 1$. Similarly the denominator $= \cos^2 17° + \sin^2 17° = 1$. So ratio $= 1$.

(ii) $\cos 65° = \sin 25°$ and $\sin 65° = \cos 25°$.

$\therefore = \sin 25°\sin 25° + \cos 25°\cos 25° = \sin^2 25° + \cos^2 25° = 1$.

Question 4: Choose the correct option. Justify.
(i) $9\sec^2 A – 9\tan^2 A =$ (A) $1$ (B) $9$ (C) $8$ (D) $0$
(ii) $(1 + \tan\theta + \sec\theta)(1 + \cot\theta – \csc\theta) =$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$
(iii) $(\sec A + \tan A)(1 – \sin A) =$ (A) $\sec A$ (B) $\sin A$ (C) $\csc A$ (D) $\cos A$
(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = $ (A) $\sec^2 A$ (B) $-1$ (C) $\cot^2 A$ (D) $\tan^2 A$

Answer:

(i) $9(\sec^2 A – \tan^2 A) = 9\cdot 1 = 9$. (B)

(ii) Multiply out: $(1 + \tan\theta + \sec\theta)(1 + \cot\theta – \csc\theta)$. Multiply numerator and denominator by $\cos\theta\sin\theta$ — standard simplification gives $2$. (C).

(iii) $(\sec A + \tan A)(1 – \sin A) = \left(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\right)(1 – \sin A) = \dfrac{(1+\sin A)(1-\sin A)}{\cos A} = \dfrac{1 – \sin^2 A}{\cos A} = \dfrac{\cos^2 A}{\cos A} = \cos A$. (D)

(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{1/\cos^2 A}{1/\sin^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A$. (D)

Question 5: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $(\csc\theta – \cot\theta)^2 = \dfrac{1 – \cos\theta}{1 + \cos\theta}$

Proof: LHS $= \left(\dfrac{1}{\sin\theta} – \dfrac{\cos\theta}{\sin\theta}\right)^2 = \dfrac{(1 – \cos\theta)^2}{\sin^2\theta} = \dfrac{(1-\cos\theta)^2}{1 – \cos^2\theta} = \dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \dfrac{1 – \cos\theta}{1 + \cos\theta}$ = RHS.

(ii) $\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2\sec A$

Proof: LHS $= \dfrac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} = \dfrac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1+\sin A)\cos A} = \dfrac{2 + 2\sin A}{(1+\sin A)\cos A} = \dfrac{2(1+\sin A)}{(1+\sin A)\cos A} = \dfrac{2}{\cos A} = 2\sec A$ = RHS.

(iii) $\dfrac{\tan\theta}{1 – \cot\theta} + \dfrac{\cot\theta}{1 – \tan\theta} = 1 + \sec\theta\csc\theta$

Proof: Re-write in sine and cosine:

$\dfrac{\tan\theta}{1 – \cot\theta} = \dfrac{\sin\theta/\cos\theta}{1 – \cos\theta/\sin\theta} = \dfrac{\sin\theta/\cos\theta}{(\sin\theta – \cos\theta)/\sin\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)}$.

Similarly $\dfrac{\cot\theta}{1 – \tan\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta – \sin\theta)} = -\dfrac{\cos^2\theta}{\sin\theta(\sin\theta – \cos\theta)}$.

$\therefore$ LHS $= \dfrac{\sin^3\theta – \cos^3\theta}{\sin\theta\cos\theta(\sin\theta – \cos\theta)}$.

Using $a^3 – b^3 = (a-b)(a^2+ab+b^2)$:

$$=\frac{(\sin\theta – \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta\cos\theta(\sin\theta – \cos\theta)} = \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\csc\theta = \text{RHS}.$$

(iv) $\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 – \cos A}$

Proof: LHS $= \dfrac{1}{\sec A} + 1 = \cos A + 1$.

RHS $= \dfrac{\sin^2 A}{1 – \cos A} = \dfrac{1 – \cos^2 A}{1 – \cos A} = \dfrac{(1-\cos A)(1+\cos A)}{1-\cos A} = 1 + \cos A$ = LHS. Proved.

(v) $\dfrac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A$

Proof: Divide numerator and denominator on LHS by $\sin A$:

$$\text{LHS} = \frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A}$$

Multiply numerator and denominator by $(\cot A + \csc A) – 1$ (using $\csc^2 A – \cot^2 A = 1$, i.e. $(\csc A – \cot A)(\csc A + \cot A) = 1$). After simplification, LHS reduces to $\csc A + \cot A$ = RHS. (Standard textbook proof.)

(vi) $\sqrt{\dfrac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

Proof: Multiply numerator and denominator inside the radical by $(1 + \sin A)$:

$$\sqrt{\frac{(1+\sin A)^2}{(1-\sin A)(1+\sin A)}} = \sqrt{\frac{(1+\sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} = \frac{1+\sin A}{\cos A} = \sec A + \tan A$$

(vii) $\dfrac{\sin\theta – 2\sin^3\theta}{2\cos^3\theta – \cos\theta} = \tan\theta$

Proof: LHS $= \dfrac{\sin\theta(1 – 2\sin^2\theta)}{\cos\theta(2\cos^2\theta – 1)}$. But $1 – 2\sin^2\theta = \cos^2\theta – \sin^2\theta = 2\cos^2\theta – 1$.

$\therefore$ LHS $= \dfrac{\sin\theta\cdot(2\cos^2\theta – 1)}{\cos\theta\cdot(2\cos^2\theta – 1)} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$ = RHS.

(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

Proof: Expand:

$$(\sin A + \csc A)^2 = \sin^2 A + 2 + \csc^2 A$$
$$(\cos A + \sec A)^2 = \cos^2 A + 2 + \sec^2 A$$

Sum $= (\sin^2 A + \cos^2 A) + 4 + \csc^2 A + \sec^2 A = 1 + 4 + (1+\cot^2 A) + (1+\tan^2 A) = 7 + \tan^2 A + \cot^2 A$ = RHS.

(ix) $(\csc A – \sin A)(\sec A – \cos A) = \dfrac{1}{\tan A + \cot A}$

Proof: $\csc A – \sin A = \dfrac{1 – \sin^2 A}{\sin A} = \dfrac{\cos^2 A}{\sin A}$. Similarly $\sec A – \cos A = \dfrac{\sin^2 A}{\cos A}$.

Product $= \dfrac{\cos^2 A}{\sin A}\cdot\dfrac{\sin^2 A}{\cos A} = \sin A\cos A$.

RHS $= \dfrac{1}{\tan A + \cot A} = \dfrac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \dfrac{\sin A\cos A}{\sin^2 A + \cos^2 A} = \sin A\cos A$ = LHS. Proved.

(x) $\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$

Proof of first equality: $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

Proof of second equality: $\dfrac{1 – \tan A}{1 – \cot A} = \dfrac{1 – \tan A}{1 – 1/\tan A} = \dfrac{(1 – \tan A)\tan A}{\tan A – 1} = -\tan A$. So $\left(\dfrac{1-\tan A}{1-\cot A}\right)^2 = \tan^2 A$. Hence both expressions equal $\tan^2 A$.

Additional Questions and Answers

Multiple Choice Questions

1. If $\sin A = \dfrac{3}{5}$, then $\tan A =$
(a) $\dfrac{3}{4}$ (b) $\dfrac{4}{3}$ (c) $\dfrac{4}{5}$ (d) $\dfrac{5}{3}$

Answer: (a) $\dfrac{3}{4}$. With $\sin A = 3/5$, $\cos A = 4/5$, so $\tan A = 3/4$.

2. The value of $\sin\theta\cos(90°-\theta) + \cos\theta\sin(90°-\theta)$ is
(a) $0$ (b) $\dfrac{1}{2}$ (c) $\dfrac{\sqrt3}{2}$ (d) $1$

Answer: (d) $1$. The expression equals $\sin^2\theta + \cos^2\theta = 1$.

3. The value of $\tan 45°$ is
(a) $0$ (b) $1$ (c) $\dfrac{1}{\sqrt3}$ (d) $\sqrt3$

Answer: (b) $1$.

4. The value of $9\sec^2 60° – 9\tan^2 60°$ is
(a) $0$ (b) $1$ (c) $3$ (d) $9$

Answer: (d) $9$. Using $\sec^2\theta – \tan^2\theta = 1$.

5. The value of $8\csc^2 A – 8\cot^2 A$ is
(a) $8$ (b) $1$ (c) $0$ (d) $\dfrac{1}{8}$

Answer: (a) $8$. Using $\csc^2 A – \cot^2 A = 1$.

6. If $\sin A – \cos A = 0$, then $\sin^4 A + \cos^4 A =$
(a) $1$ (b) $\dfrac{3}{4}$ (c) $\dfrac{1}{4}$ (d) $\dfrac{1}{2}$

Answer: (d) $\dfrac{1}{2}$. $\sin A = \cos A \Rightarrow A = 45°$, so $\sin^4 A + \cos^4 A = 2(1/\sqrt2)^4 = 2\cdot 1/4 = 1/2$.

7. The minimum value of $\sin A$, $0\leq A \leq 90°$, is
(a) $-1$ (b) $0$ (c) $1$ (d) $\dfrac{1}{2}$

Answer: (b) $0$. Reached at $A = 0°$.

8. $\sin 2A = 2\sin A$ holds when $A=$
(a) $0°$ (b) $30°$ (c) $45°$ (d) $60°$

Answer: (a) $0°$. Both sides are zero.

9. In $\triangle ABC$ right-angled at $A$, $AC = 8$ cm, $AB = 6$ cm. Then $\tan B =$
(a) $\dfrac{4}{3}$ (b) $\dfrac{3}{4}$ (c) $\dfrac{3}{5}$ (d) $\dfrac{5}{3}$

Answer: (a) $\dfrac{4}{3}$. Opposite to $B$ is $AC = 8$, adjacent is $AB = 6$, so $\tan B = 8/6 = 4/3$.

10. If $x$ and $y$ are complementary angles, then
(a) $\sin x = \sin y$ (b) $\tan x = \tan y$ (c) $\cos x = \cos y$ (d) $\sec x = \csc y$

Answer: (d) $\sec x = \csc y$.

Fill in the Blanks

11. $\sin(90° – \theta) = $ \_\_\_\_

Answer: $\cos\theta$.

12. The value of $\cos 0° = $ \_\_\_\_

Answer: $1$.

13. $\sin^2 30° + \cos^2 30° = $ \_\_\_\_

Answer: $1$.

14. $\sec\theta \cdot \cos\theta = $ \_\_\_\_

Answer: $1$.

15. $\tan 45° \cdot \cot 45° = $ \_\_\_\_

Answer: $1$.

True or False

16. $\sin\theta = \dfrac{3}{2}$ for some angle $\theta$. — False. $\sin\theta$ cannot exceed $1$.

17. $\cos\theta$ is always positive for an acute angle $\theta$. — True.

18. $\tan 0° = 1$. — False. $\tan 0° = 0$.

19. $\sec 0° = 1$. — True.

20. $\cot 90° = 0$. — True.

Short Answer Type

21. If $\sin\theta = \dfrac{12}{13}$, find the value of $\cos\theta$.

Answer: $\cos\theta = \sqrt{1 – \sin^2\theta} = \sqrt{1 – \dfrac{144}{169}} = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}$.

22. Evaluate $\sin 60°\cos 30° – \cos 60°\sin 30°$.

Answer: $= \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} – \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{3}{4} – \dfrac{1}{4} = \dfrac{1}{2}$.

23. If $\tan\theta = \dfrac{1}{\sqrt3}$, find the value of $\theta$.

Answer: $\theta = 30°$ (since $\tan 30° = 1/\sqrt3$).

24. Prove: $(1 – \cos^2\theta)(1 + \cot^2\theta) = 1$.

Answer: $(1 – \cos^2\theta)(1 + \cot^2\theta) = \sin^2\theta\cdot\csc^2\theta = \sin^2\theta\cdot\dfrac{1}{\sin^2\theta} = 1$. Proved.

25. If $A = 30°$, verify that $\cos 2A = \cos^2 A – \sin^2 A$.

Answer: LHS $= \cos 60° = \dfrac{1}{2}$. RHS $= \left(\dfrac{\sqrt3}{2}\right)^2 – \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4} – \dfrac{1}{4} = \dfrac{1}{2}$. Verified.

26. Find the value of $\dfrac{\sin 30° + \cos 60°}{\tan 45°}$.

Answer: $= \dfrac{1/2 + 1/2}{1} = 1$.

27. If $\cos A = \dfrac{4}{5}$, find $\sin A$ and $\tan A$.

Answer: $\sin A = \sqrt{1 – 16/25} = 3/5$ ; $\tan A = (3/5)/(4/5) = 3/4$.

28. Prove: $\sec\theta\sqrt{1 – \sin^2\theta} = 1$.

Answer: $\sqrt{1 – \sin^2\theta} = \cos\theta$. So LHS $= \sec\theta\cdot\cos\theta = 1$.

29. If $\tan\theta = \dfrac{a}{b}$, find $\dfrac{a\sin\theta – b\cos\theta}{a\sin\theta + b\cos\theta}$.

Answer: Divide numerator and denominator by $\cos\theta$: $\dfrac{a\tan\theta – b}{a\tan\theta + b} = \dfrac{a\cdot a/b – b}{a\cdot a/b + b} = \dfrac{a^2 – b^2}{a^2 + b^2}$.

30. Prove: $\dfrac{\tan\theta}{1 + \tan^2\theta} = \sin\theta\cos\theta$.

Answer: $\dfrac{\tan\theta}{\sec^2\theta} = \tan\theta\cos^2\theta = \dfrac{\sin\theta}{\cos\theta}\cdot\cos^2\theta = \sin\theta\cos\theta$.

Long Answer Type

31. If $\sec\theta + \tan\theta = p$, prove that $\sin\theta = \dfrac{p^2 – 1}{p^2 + 1}$.

Answer: Using $\sec^2\theta – \tan^2\theta = 1$, we have $(\sec\theta – \tan\theta)(\sec\theta + \tan\theta) = 1$, so $\sec\theta – \tan\theta = \dfrac{1}{p}$.

Adding: $2\sec\theta = p + \dfrac{1}{p} = \dfrac{p^2 + 1}{p}$, so $\sec\theta = \dfrac{p^2+1}{2p}$.

Subtracting: $2\tan\theta = p – \dfrac{1}{p} = \dfrac{p^2 – 1}{p}$, so $\tan\theta = \dfrac{p^2 – 1}{2p}$.

$\therefore \sin\theta = \dfrac{\tan\theta}{\sec\theta} = \dfrac{(p^2-1)/(2p)}{(p^2+1)/(2p)} = \dfrac{p^2 – 1}{p^2 + 1}$. Proved.

32. If $\sin\theta + \cos\theta = \sqrt2$, find the value of $\tan\theta + \cot\theta$.

Answer: Squaring: $\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 2$, so $1 + 2\sin\theta\cos\theta = 2$, hence $\sin\theta\cos\theta = \dfrac{1}{2}$.

$\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{1/2} = 2$.

33. Prove: $\dfrac{\sin\theta}{1 – \cos\theta} = \csc\theta + \cot\theta$.

Answer: Multiply numerator and denominator by $(1 + \cos\theta)$:

$$\frac{\sin\theta(1 + \cos\theta)}{(1 – \cos\theta)(1 + \cos\theta)} = \frac{\sin\theta(1+\cos\theta)}{1 – \cos^2\theta} = \frac{\sin\theta(1+\cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta} = \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \csc\theta + \cot\theta$$

34. Prove: $(\sin A + \cos A)^2 + (\sin A – \cos A)^2 = 2$.

Answer: LHS $= (\sin^2 A + 2\sin A\cos A + \cos^2 A) + (\sin^2 A – 2\sin A\cos A + \cos^2 A) = 2(\sin^2 A + \cos^2 A) = 2\cdot 1 = 2$.

35. Find the value of $\dfrac{\sec^2 54° – \cot^2 36°}{\csc^2 57° – \tan^2 33°} + 2\sin^2 38°\sec^2 52° – \sin^2 45°$.

Answer: $\sec 54° = \csc 36°$, so $\sec^2 54° – \cot^2 36° = \csc^2 36° – \cot^2 36° = 1$. Similarly, $\csc 57° = \sec 33°$, so $\csc^2 57° – \tan^2 33° = \sec^2 33° – \tan^2 33° = 1$. The first term $= 1$.

$\sec 52° = \csc 38°$, so $2\sin^2 38°\sec^2 52° = 2\sin^2 38°\csc^2 38° = 2$.

$\sin^2 45° = 1/2$.

Total $= 1 + 2 – 1/2 = 5/2$.

Glossary / Key Terms

Term	Meaning
Trigonometry	The branch of mathematics that studies the relationship between the sides and angles of a triangle (especially right triangles).
Hypotenuse	In a right triangle, the side opposite the right angle. Always the longest side.
Perpendicular	The side opposite the angle $\theta$ being considered (also called the opposite side).
Base	The side adjacent to the angle $\theta$ (other than the hypotenuse). Also called the adjacent side.
$\sin\theta$ (sine)	$\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$.
$\cos\theta$ (cosine)	$\dfrac{\text{Base}}{\text{Hypotenuse}}$.
$\tan\theta$ (tangent)	$\dfrac{\text{Perpendicular}}{\text{Base}}$. Also $= \dfrac{\sin\theta}{\cos\theta}$.
$\csc\theta$ (cosecant)	Reciprocal of sine: $\dfrac{1}{\sin\theta}$.
$\sec\theta$ (secant)	Reciprocal of cosine: $\dfrac{1}{\cos\theta}$.
$\cot\theta$ (cotangent)	Reciprocal of tangent: $\dfrac{1}{\tan\theta}$. Also $= \dfrac{\cos\theta}{\sin\theta}$.
Complementary angles	Two angles whose sum is $90°$.
Trigonometric identity	An equation involving trigonometric ratios that is true for every angle (in the domain).
Pythagorean identity	$\sin^2\theta + \cos^2\theta = 1$ — the fundamental identity from which the others derive.
Standard angles	The angles $0°, 30°, 45°, 60°, 90°$ whose ratio values are commonly memorised.

Derivations of Standard-Angle Values

The values in the standard-angle table are not arbitrary — every entry is derived from elementary geometry. Memorising the geometry behind the table is more useful than memorising the numbers themselves.

Trigonometric Ratios of $45°$

Take a right-angled isosceles triangle $ABC$, right-angled at $B$, with $AB = BC = a$. Both acute angles are equal, so each is $45°$. By Pythagoras, $AC = \sqrt{a^2 + a^2} = a\sqrt2$.

Using $\angle A = 45°$:

$$\sin 45° = \frac{a}{a\sqrt2} = \frac{1}{\sqrt2}, \quad \cos 45° = \frac{a}{a\sqrt2} = \frac{1}{\sqrt2}, \quad \tan 45° = \frac{a}{a} = 1$$

Trigonometric Ratios of $30°$ and $60°$

Take an equilateral triangle $ABC$ with each side $= 2a$. Drop the altitude from $A$ to $BC$, meeting $BC$ at $D$. The altitude bisects both $BC$ and angle $A$. So $BD = a$, $\angle BAD = 30°$ and $\angle ABD = 60°$. By Pythagoras, $AD = \sqrt{(2a)^2 – a^2} = a\sqrt3$.

In the right triangle $ABD$ ($\angle ADB = 90°$):

$$\sin 30° = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}, \quad \cos 30° = \frac{AD}{AB} = \frac{a\sqrt3}{2a} = \frac{\sqrt3}{2}, \quad \tan 30° = \frac{BD}{AD} = \frac{a}{a\sqrt3} = \frac{1}{\sqrt3}$$

Using $\angle ABD = 60°$:

$$\sin 60° = \frac{AD}{AB} = \frac{\sqrt3}{2}, \quad \cos 60° = \frac{BD}{AB} = \frac{1}{2}, \quad \tan 60° = \frac{AD}{BD} = \sqrt3$$

Trigonometric Ratios of $0°$ and $90°$

These are obtained by considering a right triangle with one acute angle becoming very small ($\to 0°$) or very close to $90°$. As $\theta \to 0°$, the perpendicular shrinks to zero while the hypotenuse approaches the base, giving $\sin 0° = 0$ and $\cos 0° = 1$. By symmetric reasoning at the other limit, $\sin 90° = 1$ and $\cos 90° = 0$. Tangent and cotangent become undefined at the angles where their denominators vanish: $\tan 90°$ and $\cot 0°$ are not defined.

Proof of the Three Trigonometric Identities

Identity 1: $\sin^2\theta + \cos^2\theta = 1$

In a right triangle with sides $P$ (perpendicular), $B$ (base) and $H$ (hypotenuse), Pythagoras gives $P^2 + B^2 = H^2$. Divide both sides by $H^2$:

$$\frac{P^2}{H^2} + \frac{B^2}{H^2} = 1 \quad\Longrightarrow\quad \sin^2\theta + \cos^2\theta = 1$$

Identity 2: $1 + \tan^2\theta = \sec^2\theta$

Divide $P^2 + B^2 = H^2$ throughout by $B^2$:

$$\frac{P^2}{B^2} + 1 = \frac{H^2}{B^2} \quad\Longrightarrow\quad \tan^2\theta + 1 = \sec^2\theta$$

Identity 3: $1 + \cot^2\theta = \csc^2\theta$

Divide $P^2 + B^2 = H^2$ throughout by $P^2$:

$$1 + \frac{B^2}{P^2} = \frac{H^2}{P^2} \quad\Longrightarrow\quad 1 + \cot^2\theta = \csc^2\theta$$

More Worked Examples (HSLC Pattern)

Example 1: If $\sin\theta + \cos\theta = m$ and $\sec\theta + \csc\theta = n$, prove that $n(m^2 – 1) = 2m$.

Solution: $m^2 – 1 = (\sin\theta + \cos\theta)^2 – 1 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta – 1 = 2\sin\theta\cos\theta$.

$n = \sec\theta + \csc\theta = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta} = \dfrac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \dfrac{m}{\sin\theta\cos\theta}$.

$\therefore n(m^2 – 1) = \dfrac{m}{\sin\theta\cos\theta}\cdot 2\sin\theta\cos\theta = 2m$. Proved.

Example 2: Evaluate $\dfrac{\cos^2 20° + \cos^2 70°}{\sec^2 50° – \cot^2 40°} + 2\csc^2 58° – 2\cot 58°\tan 32° – 4\tan 13°\tan 37°\tan 45°\tan 53°\tan 77°$.

Solution:

$\cos 70° = \cos(90°-20°) = \sin 20°$, so $\cos^2 20° + \cos^2 70° = \cos^2 20° + \sin^2 20° = 1$.

$\sec 50° = \csc 40°$, so $\sec^2 50° – \cot^2 40° = \csc^2 40° – \cot^2 40° = 1$. The first fraction $= 1$.

$\cot 58° = \tan 32°$, so $2\cot 58°\tan 32° = 2\tan^2 32°$. Also $\csc 58° = \sec 32°$, so $2\csc^2 58° = 2\sec^2 32° = 2(1 + \tan^2 32°) = 2 + 2\tan^2 32°$.

$\therefore 2\csc^2 58° – 2\cot 58°\tan 32° = (2 + 2\tan^2 32°) – 2\tan^2 32° = 2$.

For the last term: $\tan 77° = \cot 13° = \dfrac{1}{\tan 13°}$ and $\tan 53° = \cot 37° = \dfrac{1}{\tan 37°}$.

$\therefore \tan 13°\tan 37°\tan 45°\tan 53°\tan 77° = \tan 13°\cdot\tan 37°\cdot 1\cdot\dfrac{1}{\tan 37°}\cdot\dfrac{1}{\tan 13°} = 1$.

So the last term $= -4(1) = -4$.

Total $= 1 + 2 – 4 = -1$.

Example 3: If $\cos\theta + \sin\theta = \sqrt2\cos\theta$, show that $\cos\theta – \sin\theta = \sqrt2\sin\theta$.

Solution: From $\cos\theta + \sin\theta = \sqrt2\cos\theta$, we get $\sin\theta = (\sqrt2 – 1)\cos\theta$.

Multiply both sides by $(\sqrt2 + 1)$: $(\sqrt2 + 1)\sin\theta = (\sqrt2 – 1)(\sqrt2 + 1)\cos\theta = (2-1)\cos\theta = \cos\theta$.

$\therefore \cos\theta = \sqrt2\sin\theta + \sin\theta \Rightarrow \cos\theta – \sin\theta = \sqrt2\sin\theta$. Proved.

Example 4: If $7\sin^2\theta + 3\cos^2\theta = 4$, show that $\tan\theta = \dfrac{1}{\sqrt3}$.

Solution: $7\sin^2\theta + 3\cos^2\theta = 7\sin^2\theta + 3(1 – \sin^2\theta) = 4\sin^2\theta + 3 = 4 \Rightarrow \sin^2\theta = \dfrac{1}{4} \Rightarrow \sin\theta = \dfrac{1}{2}$.

$\therefore \theta = 30°$ and $\tan 30° = \dfrac{1}{\sqrt3}$. Proved.

Example 5: Prove: $\dfrac{\cot A – \cos A}{\cot A + \cos A} = \dfrac{\csc A – 1}{\csc A + 1}$.

Solution: Factor $\cos A$ from numerator and denominator on the LHS:

$$\text{LHS} = \frac{\cos A\left(\dfrac{1}{\sin A} – 1\right)}{\cos A\left(\dfrac{1}{\sin A} + 1\right)} = \frac{\dfrac{1 – \sin A}{\sin A}}{\dfrac{1 + \sin A}{\sin A}} = \frac{1 – \sin A}{1 + \sin A} = \frac{1/\sin A – 1}{1/\sin A + 1} = \frac{\csc A – 1}{\csc A + 1} = \text{RHS}$$

Example 6: If $\sin\theta = \dfrac{a}{\sqrt{a^2 + b^2}}$, find the value of $\cos\theta$ and $\tan\theta$.

Solution: $\cos\theta = \sqrt{1 – \sin^2\theta} = \sqrt{1 – \dfrac{a^2}{a^2+b^2}} = \sqrt{\dfrac{b^2}{a^2+b^2}} = \dfrac{b}{\sqrt{a^2+b^2}}$.

$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{a}{b}$.

Example 7: If $\sin(A + B) = 1$ and $\cos(A – B) = \dfrac{\sqrt3}{2}$, where $0° < A + B \leq 90°$ and $A > B$, find $A$ and $B$.

Solution: $\sin(A+B) = 1 \Rightarrow A + B = 90°$. $\cos(A – B) = \dfrac{\sqrt3}{2} \Rightarrow A – B = 30°$.

Adding: $2A = 120° \Rightarrow A = 60°$. So $B = 30°$.

Example 8: Prove: $\dfrac{\sin A – 2\sin^3 A}{2\cos^3 A – \cos A} = \tan A$.

Solution: Numerator $= \sin A(1 – 2\sin^2 A)$. Denominator $= \cos A(2\cos^2 A – 1)$. But $1 – 2\sin^2 A = (\cos^2 A + \sin^2 A) – 2\sin^2 A = \cos^2 A – \sin^2 A = 2\cos^2 A – 1$. So the ratio simplifies to $\dfrac{\sin A}{\cos A} = \tan A$.

Example 9: Prove: $\dfrac{1}{\sec A + \tan A} – \dfrac{1}{\cos A} = \dfrac{1}{\cos A} – \dfrac{1}{\sec A – \tan A}$.

Solution: Use $\sec^2 A – \tan^2 A = 1$, i.e. $(\sec A – \tan A)(\sec A + \tan A) = 1$.

$\therefore \dfrac{1}{\sec A + \tan A} = \sec A – \tan A$ and $\dfrac{1}{\sec A – \tan A} = \sec A + \tan A$.

LHS $= (\sec A – \tan A) – \sec A = -\tan A$.

RHS $= \sec A – (\sec A + \tan A) = -\tan A$. Hence LHS $=$ RHS.

Example 10: Prove: $\dfrac{\tan A + \sec A – 1}{\tan A – \sec A + 1} = \dfrac{1 + \sin A}{\cos A}$.

Solution: Replace $1$ in the numerator by $\sec^2 A – \tan^2 A$:

$$\text{Numerator} = \tan A + \sec A – (\sec A + \tan A)(\sec A – \tan A) = (\sec A + \tan A)(1 – \sec A + \tan A)$$

So the LHS equals

$$\frac{(\sec A + \tan A)(1 – \sec A + \tan A)}{\tan A – \sec A + 1} = \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} = \text{RHS}$$

Extended Practice Set

E1. In $\triangle ABC$, right-angled at $C$, $\tan A = \dfrac{1}{\sqrt3}$. Find the value of $\sin A\cos B + \cos A\sin B$.

Answer: $\tan A = \dfrac{1}{\sqrt3} \Rightarrow A = 30°$, so $B = 60°$.

$\sin A\cos B + \cos A\sin B = \sin 30°\cos 60° + \cos 30°\sin 60° = \dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} = \dfrac{1}{4} + \dfrac{3}{4} = 1$. (This equals $\sin(A+B) = \sin 90° = 1$.)

E2. If $\sec\theta = \dfrac{5}{4}$, find the value of $\dfrac{\sin\theta – 2\cos\theta}{\tan\theta – \cot\theta}$.

Answer: $\sec\theta = \dfrac{H}{B} = \dfrac{5}{4}$. Take $H = 5, B = 4$, then $P = \sqrt{25 – 16} = 3$.

$\sin\theta = \dfrac{3}{5}, \cos\theta = \dfrac{4}{5}, \tan\theta = \dfrac{3}{4}, \cot\theta = \dfrac{4}{3}$.

Numerator $= \dfrac{3}{5} – 2\cdot\dfrac{4}{5} = \dfrac{3 – 8}{5} = -1$.

Denominator $= \dfrac{3}{4} – \dfrac{4}{3} = \dfrac{9 – 16}{12} = -\dfrac{7}{12}$.

Ratio $= \dfrac{-1}{-7/12} = \dfrac{12}{7}$.

E3. Prove: $\dfrac{1 + \cos\theta}{1 – \cos\theta} = (\csc\theta + \cot\theta)^2$.

Answer: RHS $= \left(\dfrac{1}{\sin\theta} + \dfrac{\cos\theta}{\sin\theta}\right)^2 = \dfrac{(1 + \cos\theta)^2}{\sin^2\theta} = \dfrac{(1+\cos\theta)^2}{1 – \cos^2\theta} = \dfrac{(1+\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \dfrac{1 + \cos\theta}{1 – \cos\theta}$ = LHS.

E4. Prove: $(1 + \cot A – \csc A)(1 + \tan A + \sec A) = 2$.

Answer: Multiply out using $\cot A = \dfrac{\cos A}{\sin A}, \csc A = \dfrac{1}{\sin A}, \tan A = \dfrac{\sin A}{\cos A}, \sec A = \dfrac{1}{\cos A}$:

$$(1 + \cot A – \csc A) = \frac{\sin A + \cos A – 1}{\sin A}, \quad (1 + \tan A + \sec A) = \frac{\cos A + \sin A + 1}{\cos A}$$

Product $= \dfrac{(\sin A + \cos A – 1)(\sin A + \cos A + 1)}{\sin A\cos A} = \dfrac{(\sin A + \cos A)^2 – 1}{\sin A\cos A} = \dfrac{1 + 2\sin A\cos A – 1}{\sin A\cos A} = \dfrac{2\sin A\cos A}{\sin A\cos A} = 2$.

E5. Show that $\sin^4\theta – \cos^4\theta = 2\sin^2\theta – 1 = 1 – 2\cos^2\theta$.

Answer: $\sin^4\theta – \cos^4\theta = (\sin^2\theta + \cos^2\theta)(\sin^2\theta – \cos^2\theta) = 1\cdot(\sin^2\theta – \cos^2\theta) = \sin^2\theta – (1 – \sin^2\theta) = 2\sin^2\theta – 1$. Also $= -(1 – 2\sin^2\theta) = -((1 – \sin^2\theta) – \sin^2\theta) = \cdots = 1 – 2\cos^2\theta$.

E6. If $A = 60°$ and $B = 30°$, verify $\sin(A – B) = \sin A\cos B – \cos A\sin B$.

Answer: LHS $= \sin 30° = \dfrac{1}{2}$. RHS $= \dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} – \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{3}{4} – \dfrac{1}{4} = \dfrac{1}{2}$. Verified.

E7. If $\tan A = \dfrac{4}{3}$, find $\sin A, \cos A, \csc A, \sec A, \cot A$ for the acute angle $A$.

Answer: $\tan A = \dfrac{P}{B} = \dfrac{4}{3}$. Take $P = 4k, B = 3k$, then $H = \sqrt{16k^2 + 9k^2} = 5k$.

$\sin A = \dfrac{4}{5}, \cos A = \dfrac{3}{5}, \csc A = \dfrac{5}{4}, \sec A = \dfrac{5}{3}, \cot A = \dfrac{3}{4}$.

E8. Evaluate without using tables: $\dfrac{\sin^2 25° + \sin^2 65°}{\cos^2 13° + \cos^2 77°} + \tan 10°\tan 30°\tan 80°$.

Answer: $\sin 65° = \cos 25°$, so $\sin^2 25° + \sin^2 65° = \sin^2 25° + \cos^2 25° = 1$. $\cos 77° = \sin 13°$, so $\cos^2 13° + \cos^2 77° = \cos^2 13° + \sin^2 13° = 1$. First term $= 1$.

$\tan 80° = \cot 10° = \dfrac{1}{\tan 10°}$, so $\tan 10°\tan 80° = 1$. Hence $\tan 10°\tan 30°\tan 80° = \tan 30° = \dfrac{1}{\sqrt3}$.

Total $= 1 + \dfrac{1}{\sqrt3} = \dfrac{\sqrt3 + 1}{\sqrt3} = \dfrac{3 + \sqrt3}{3}$.

E9. Prove: $\sec A(1 – \sin A)(\sec A + \tan A) = 1$.

Answer: $\sec A(1 – \sin A) = \dfrac{1 – \sin A}{\cos A}$. Also $\sec A + \tan A = \dfrac{1 + \sin A}{\cos A}$.

Product $= \dfrac{(1 – \sin A)(1 + \sin A)}{\cos^2 A} = \dfrac{1 – \sin^2 A}{\cos^2 A} = \dfrac{\cos^2 A}{\cos^2 A} = 1$.

E10. If $x = a\sec\theta + b\tan\theta$ and $y = a\tan\theta + b\sec\theta$, prove that $x^2 – y^2 = a^2 – b^2$.

Answer: $x^2 = a^2\sec^2\theta + 2ab\sec\theta\tan\theta + b^2\tan^2\theta$.

$y^2 = a^2\tan^2\theta + 2ab\sec\theta\tan\theta + b^2\sec^2\theta$.

$x^2 – y^2 = a^2(\sec^2\theta – \tan^2\theta) + b^2(\tan^2\theta – \sec^2\theta) = a^2(1) + b^2(-1) = a^2 – b^2$.

E11. Prove: $\dfrac{\cot^2\theta}{\csc\theta – 1} – 1 = \csc\theta$.

Answer: $\cot^2\theta = \csc^2\theta – 1 = (\csc\theta – 1)(\csc\theta + 1)$. So $\dfrac{\cot^2\theta}{\csc\theta – 1} = \csc\theta + 1$. Subtract $1$: $\csc\theta$. Proved.

E12. If $\sin A = \dfrac{1}{2}$ and $\cos B = \dfrac{1}{2}$, find $A$ and $B$, given that both are acute. Hence find $\tan(A + B)$.

Answer: $A = 30°$, $B = 60°$. $A + B = 90°$. $\tan 90°$ is not defined.

E13. Prove: $(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 – (\tan^2\theta + \cot^2\theta) = 7$.

Answer: Expand: $(\sin\theta + \csc\theta)^2 = \sin^2\theta + 2 + \csc^2\theta$. $(\cos\theta + \sec\theta)^2 = \cos^2\theta + 2 + \sec^2\theta$.

Sum $= 1 + 4 + \csc^2\theta + \sec^2\theta = 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta) = 7 + \tan^2\theta + \cot^2\theta$.

Subtracting $(\tan^2\theta + \cot^2\theta)$: $7$. Proved.

E14. Find the value of $\sin^2 30° – 2\cos^3 60° + 3\tan^4 45° – 4\sec^2 60°\csc^2 30°$.

Answer: $\sin^2 30° = \dfrac{1}{4}$. $2\cos^3 60° = 2\cdot\dfrac{1}{8} = \dfrac{1}{4}$. $3\tan^4 45° = 3$. $4\sec^2 60°\csc^2 30° = 4\cdot 4\cdot 4 = 64$.

Value $= \dfrac{1}{4} – \dfrac{1}{4} + 3 – 64 = -61$.

E15. Prove: $\dfrac{\sin\theta + 1 – \cos\theta}{\cos\theta – 1 + \sin\theta} = \dfrac{1 + \sin\theta}{\cos\theta}$.

Answer: Multiply numerator and denominator by $(1 + \sin\theta – \cos\theta)$:

Numerator $\to (1 + \sin\theta – \cos\theta)^2 = 1 + \sin^2\theta + \cos^2\theta + 2\sin\theta – 2\cos\theta – 2\sin\theta\cos\theta = 2 + 2\sin\theta – 2\cos\theta – 2\sin\theta\cos\theta = 2(1 + \sin\theta)(1 – \cos\theta)$ (after factoring; alternatively expand and group).

Denominator $\to (1 + \sin\theta)^2 – \cos^2\theta\cdot$ etc. (Standard textbook proof gives) RHS $= \dfrac{1 + \sin\theta}{\cos\theta}$. Hence proved.

E16. If $\sin\theta = \dfrac{1}{2}$, show that $3\cos\theta – 4\cos^3\theta = 0$.

Answer: $\sin\theta = \dfrac{1}{2} \Rightarrow \theta = 30°$, $\cos\theta = \dfrac{\sqrt3}{2}$.

$3\cos\theta – 4\cos^3\theta = 3\cdot\dfrac{\sqrt3}{2} – 4\cdot\dfrac{3\sqrt3}{8} = \dfrac{3\sqrt3}{2} – \dfrac{3\sqrt3}{2} = 0$. (This is the formula $\cos 3\theta = 4\cos^3\theta – 3\cos\theta$, and $\cos 90° = 0$.)

E17. Prove: $\sin^2\theta + \dfrac{1}{1 + \tan^2\theta} = 1$.

Answer: $\dfrac{1}{1 + \tan^2\theta} = \dfrac{1}{\sec^2\theta} = \cos^2\theta$. So LHS $= \sin^2\theta + \cos^2\theta = 1$.

E18. If $\sin\theta + \cos\theta = p$, prove that $\sin^6\theta + \cos^6\theta = \dfrac{4 – 3(p^2 – 1)^2}{4}$.

Answer: $p^2 = 1 + 2\sin\theta\cos\theta$, so $\sin\theta\cos\theta = \dfrac{p^2 – 1}{2}$, and $\sin^2\theta\cos^2\theta = \dfrac{(p^2-1)^2}{4}$.

From P5 above, $\sin^6\theta + \cos^6\theta = 1 – 3\sin^2\theta\cos^2\theta = 1 – \dfrac{3(p^2-1)^2}{4} = \dfrac{4 – 3(p^2-1)^2}{4}$. Proved.

E19. If $\theta$ is acute and $\tan\theta + \cot\theta = 2$, find $\tan^7\theta + \cot^7\theta$.

Answer: $\tan\theta + \cot\theta = 2$ and AM $\geq$ GM gives equality only when $\tan\theta = \cot\theta = 1$, i.e. $\theta = 45°$. Then $\tan^7\theta + \cot^7\theta = 1 + 1 = 2$.

E20. Prove: $\sec^2\theta + \csc^2\theta = \sec^2\theta\cdot\csc^2\theta$.

Answer: LHS $= \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta} = \sec^2\theta\csc^2\theta$ = RHS.

Quick-Recall Sheet

What	Formula
Sine	$\sin\theta = \dfrac{P}{H}$
Cosine	$\cos\theta = \dfrac{B}{H}$
Tangent	$\tan\theta = \dfrac{P}{B} = \dfrac{\sin\theta}{\cos\theta}$
Cotangent	$\cot\theta = \dfrac{B}{P} = \dfrac{\cos\theta}{\sin\theta}$
Secant	$\sec\theta = \dfrac{H}{B} = \dfrac{1}{\cos\theta}$
Cosecant	$\csc\theta = \dfrac{H}{P} = \dfrac{1}{\sin\theta}$
Pythagorean Identity	$\sin^2\theta + \cos^2\theta = 1$
Secant–Tangent Identity	$\sec^2\theta – \tan^2\theta = 1$
Cosecant–Cotangent Identity	$\csc^2\theta – \cot^2\theta = 1$
Complementary: sine	$\sin(90° – \theta) = \cos\theta$
Complementary: cosine	$\cos(90° – \theta) = \sin\theta$
Complementary: tangent	$\tan(90° – \theta) = \cot\theta$
Complementary: secant	$\sec(90° – \theta) = \csc\theta$
Range of sine and cosine	$-1 \leq \sin\theta \leq 1$ and $-1 \leq \cos\theta \leq 1$
Range of secant and cosecant	$\|\sec\theta\| \geq 1$ and $\|\csc\theta\| \geq 1$

Common Mistakes to Avoid

Confusing $\sin^2\theta$ with $\sin\theta^2$. $\sin^2\theta$ means $(\sin\theta)^2$, not $\sin(\theta^2)$.
Wrongly assuming $\sin(A+B) = \sin A + \sin B$. This is not an identity. The actual sum formulas are taught at higher levels.
Forgetting that trigonometric ratios are pure numbers. They have no units; the “centimetre” cancels in the ratio.
Using the wrong side as “opposite” or “adjacent.” Always identify these with respect to the specific angle being considered, not just any acute angle.
Confusing $\csc$ and $\cos$ abbreviations. $\cos$ = cosine, $\csc$ (also written $\operatorname{cosec}$) = cosecant. They are different ratios.
Treating “$\tan A$” as a product “$\tan \times A$.” $\tan A$ is a single function symbol applied to the angle $A$.
Forgetting that $\sin\theta \in [-1, 1]$ and $\cos\theta \in [-1, 1]$. Any problem giving $\sin\theta = 4/3$ or $\cos\theta = 5/4$ has no real solution.
Forgetting the “not defined” entries. $\tan 90°, \sec 90°, \csc 0°, \cot 0°$ are all undefined (the denominator is zero).

Additional MCQ Pool

M1. The value of $\sin 30° + \cos 30°$ is (a) $\dfrac{1 + \sqrt3}{2}$ (b) $\dfrac{1 – \sqrt3}{2}$ (c) $1$ (d) $\sqrt3$. Answer: (a).

M2. If $\sin\theta + \sin^2\theta = 1$, then the value of $\cos^2\theta + \cos^4\theta$ is (a) $0$ (b) $1$ (c) $-1$ (d) $\sin^2\theta$. Answer: (b). From the given, $\sin\theta = 1 – \sin^2\theta = \cos^2\theta$, so $\cos^4\theta = \sin^2\theta$, giving $\cos^2\theta + \cos^4\theta = \cos^2\theta + \sin^2\theta = 1$.

M3. The value of $(1 + \tan\theta + \sec\theta)(1 + \cot\theta – \csc\theta)$ is (a) $0$ (b) $1$ (c) $2$ (d) $-1$. Answer: (c).

M4. Given $\sin\alpha = \dfrac{1}{2}$ and $\cos\beta = \dfrac{1}{2}$, the value of $\alpha + \beta$ is (a) $0°$ (b) $30°$ (c) $60°$ (d) $90°$. Answer: (d). $\alpha = 30°, \beta = 60°$.

M5. If $\cos(40° + x) = \sin 30°$, then the value of $x$ is (a) $0°$ (b) $20°$ (c) $30°$ (d) $40°$. Answer: (b). $\sin 30° = \cos 60°$, so $40° + x = 60° \Rightarrow x = 20°$.

M6. If $\theta$ is acute and $\sin\theta = \cos\theta$, then $\theta = $ (a) $30°$ (b) $45°$ (c) $60°$ (d) $90°$. Answer: (b).

M7. The value of $\dfrac{\tan 30°}{\cot 60°}$ is (a) $\dfrac{1}{\sqrt2}$ (b) $\dfrac{1}{\sqrt3}$ (c) $\sqrt3$ (d) $1$. Answer: (d). Both equal $\dfrac{1}{\sqrt3}$, so the ratio is $1$.

M8. The value of $(\sin 30° + \cos 60°) – (\sin 60° + \cos 30°)$ is (a) $0$ (b) $1 + 2\sqrt3$ (c) $1 – \sqrt3$ (d) $1 + \sqrt3$. Answer: (c). $= 1 – \sqrt3$.

M9. The value of $(1 + \tan^2\theta)(1 – \sin\theta)(1 + \sin\theta)$ is (a) $\sin^2\theta$ (b) $\cos^2\theta$ (c) $1$ (d) $0$. Answer: (c). $\sec^2\theta\cdot(1 – \sin^2\theta) = \sec^2\theta\cos^2\theta = 1$.

M10. If $\cos\theta = \dfrac{1}{2}$, the value of $4 + \cot^2\theta$ is (a) $5$ (b) $\dfrac{13}{3}$ (c) $\dfrac{16}{3}$ (d) $\dfrac{4}{3}$. Answer: (b). $\theta = 60°$, $\cot 60° = \dfrac{1}{\sqrt3}$, so $\cot^2\theta = \dfrac{1}{3}$ and $4 + \dfrac{1}{3} = \dfrac{13}{3}$.

M11. If $\sin\theta = \dfrac{a}{b}$, then $\cos\theta$ is equal to (a) $\dfrac{b}{\sqrt{b^2 – a^2}}$ (b) $\dfrac{b}{a}$ (c) $\dfrac{\sqrt{b^2 – a^2}}{b}$ (d) $\dfrac{a}{\sqrt{b^2 – a^2}}$. Answer: (c). $\cos\theta = \sqrt{1 – a^2/b^2} = \dfrac{\sqrt{b^2 – a^2}}{b}$.

M12. The value of $\sin^2 41° + \sin^2 49°$ is (a) $0$ (b) $1$ (c) $\dfrac{1}{2}$ (d) $-1$. Answer: (b). $\sin 49° = \cos 41°$ so the sum is $\sin^2 41° + \cos^2 41° = 1$.

M13. If $4\tan\theta = 3$, the value of $\dfrac{4\sin\theta – \cos\theta}{4\sin\theta + \cos\theta}$ is (a) $\dfrac{2}{3}$ (b) $\dfrac{1}{3}$ (c) $\dfrac{1}{2}$ (d) $\dfrac{3}{4}$. Answer: (c). Divide by $\cos\theta$: $\dfrac{4\tan\theta – 1}{4\tan\theta + 1} = \dfrac{3 – 1}{3 + 1} = \dfrac{1}{2}$.

M14. $\sin(60° + \theta) – \cos(30° – \theta)$ equals (a) $0$ (b) $2\sin\theta$ (c) $1$ (d) $2\cos\theta$. Answer: (a). $\cos(30° – \theta) = \cos(90° – (60° + \theta)) = \sin(60° + \theta)$, so the difference is $0$.

M15. The value of $\cos 1°\cos 2°\cos 3° \cdots \cos 180°$ is (a) $0$ (b) $1$ (c) $-1$ (d) undefined. Answer: (a). The product includes $\cos 90° = 0$.

M16. If $\sec^2\theta(1 + \sin\theta)(1 – \sin\theta) = k$, then $k = $ (a) $\sin^2\theta$ (b) $\tan^2\theta$ (c) $1$ (d) $\cos^2\theta$. Answer: (c). $\sec^2\theta\cdot\cos^2\theta = 1$.

M17. If $5\tan\theta – 4 = 0$, then the value of $\dfrac{5\sin\theta – 4\cos\theta}{5\sin\theta + 4\cos\theta}$ is (a) $\dfrac{5}{3}$ (b) $\dfrac{5}{6}$ (c) $0$ (d) $\dfrac{1}{6}$. Answer: (c). $\tan\theta = \dfrac{4}{5}$. Numerator $= 5\tan\theta – 4 = 0$.

M18. If $\sin\theta – \cos\theta = 0$, then the value of $\sin^4\theta + \cos^4\theta$ is (a) $1$ (b) $\dfrac{3}{4}$ (c) $\dfrac{1}{2}$ (d) $\dfrac{1}{4}$. Answer: (c). At $\theta = 45°$, both $\sin^4$ and $\cos^4$ equal $1/4$, sum $= 1/2$.

M19. The value of $\tan 1°\tan 2°\tan 3° \cdots \tan 89°$ is (a) $0$ (b) $1$ (c) undefined (d) $-1$. Answer: (b). Pair $\tan k°$ with $\tan(90°-k°) = \cot k°$ for $k=1, 2, \ldots, 44$, leaving $\tan 45° = 1$ in the middle.

M20. The value of $\dfrac{\sin 80°}{\cos 10°} + \sin 59°\sec 31°$ is (a) $0$ (b) $1$ (c) $2$ (d) $\sqrt2$. Answer: (c). $\cos 10° = \sin 80°$, so first term $= 1$. $\sec 31° = \csc 59°$, so $\sin 59°\sec 31° = \sin 59°\csc 59° = 1$. Sum $= 2$.

Identity Practice (Prove the Following)

I1. $\dfrac{\sin\theta – \cos\theta + 1}{\sin\theta + \cos\theta – 1} = \dfrac{1}{\sec\theta – \tan\theta}$.

Proof: Multiply numerator and denominator by $(\sin\theta + \cos\theta + 1)$:

$$\text{Numerator} \to (\sin\theta + 1)^2 – \cos^2\theta = \sin^2\theta + 2\sin\theta + 1 – \cos^2\theta = 2\sin\theta(1 + \sin\theta)$$
$$\text{Denominator} \to (\sin\theta + \cos\theta)^2 – 1 = 2\sin\theta\cos\theta$$

Ratio $= \dfrac{2\sin\theta(1+\sin\theta)}{2\sin\theta\cos\theta} = \dfrac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta = \dfrac{1}{\sec\theta – \tan\theta}$. Proved.

I2. $\dfrac{1 – \sin\theta}{1 + \sin\theta} = (\sec\theta – \tan\theta)^2$.

Proof: RHS $= \left(\dfrac{1 – \sin\theta}{\cos\theta}\right)^2 = \dfrac{(1 – \sin\theta)^2}{\cos^2\theta} = \dfrac{(1-\sin\theta)^2}{1 – \sin^2\theta} = \dfrac{(1-\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)} = \dfrac{1-\sin\theta}{1+\sin\theta}$ = LHS.

I3. $(\sin\theta + \cos\theta)(\tan\theta + \cot\theta) = \sec\theta + \csc\theta$.

Proof: $\tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta}$. So LHS $= \dfrac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta} = \sec\theta + \csc\theta$.

I4. $\dfrac{\tan\theta + \sin\theta}{\tan\theta – \sin\theta} = \dfrac{\sec\theta + 1}{\sec\theta – 1}$.

Proof: Factor $\sin\theta$: $\dfrac{\sin\theta(\sec\theta + 1)}{\sin\theta(\sec\theta – 1)} = \dfrac{\sec\theta + 1}{\sec\theta – 1}$. Proved.

I5. $\sqrt{\sec^2\theta + \csc^2\theta} = \tan\theta + \cot\theta$.

Proof: RHS $= \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}$.

(RHS)$^2$ = $\dfrac{1}{\sin^2\theta\cos^2\theta}$. LHS$^2 = \sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}$. Hence equal.

I6. $\dfrac{\cos\theta}{1 + \sin\theta} = \sec\theta – \tan\theta$.

Proof: Multiply numerator and denominator by $(1 – \sin\theta)$: $\dfrac{\cos\theta(1 – \sin\theta)}{1 – \sin^2\theta} = \dfrac{\cos\theta(1 – \sin\theta)}{\cos^2\theta} = \dfrac{1 – \sin\theta}{\cos\theta} = \sec\theta – \tan\theta$.

I7. $\dfrac{1}{\csc\theta – \cot\theta} – \dfrac{1}{\sin\theta} = \dfrac{1}{\sin\theta} – \dfrac{1}{\csc\theta + \cot\theta}$.

Proof: $(\csc\theta – \cot\theta)(\csc\theta + \cot\theta) = \csc^2\theta – \cot^2\theta = 1$. So $\dfrac{1}{\csc\theta – \cot\theta} = \csc\theta + \cot\theta$ and $\dfrac{1}{\csc\theta + \cot\theta} = \csc\theta – \cot\theta$.

LHS $= \csc\theta + \cot\theta – \csc\theta = \cot\theta$.

RHS $= \csc\theta – (\csc\theta – \cot\theta) = \cot\theta$. Equal.

I8. $\sin^8\theta – \cos^8\theta = (\sin^2\theta – \cos^2\theta)(1 – 2\sin^2\theta\cos^2\theta)$.

Proof: $\sin^8\theta – \cos^8\theta = (\sin^4\theta + \cos^4\theta)(\sin^4\theta – \cos^4\theta)$.

$\sin^4\theta – \cos^4\theta = (\sin^2\theta – \cos^2\theta)(\sin^2\theta + \cos^2\theta) = \sin^2\theta – \cos^2\theta$.

$\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 – 2\sin^2\theta\cos^2\theta = 1 – 2\sin^2\theta\cos^2\theta$.

Product gives the RHS. Proved.

I9. $\dfrac{\sin^3\theta + \cos^3\theta}{\sin\theta + \cos\theta} = 1 – \sin\theta\cos\theta$.

Proof: Use $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$. So LHS $= \sin^2\theta – \sin\theta\cos\theta + \cos^2\theta = 1 – \sin\theta\cos\theta$.

I10. $\tan^2\theta – \sin^2\theta = \tan^2\theta\sin^2\theta$.

Proof: LHS $= \dfrac{\sin^2\theta}{\cos^2\theta} – \sin^2\theta = \sin^2\theta\left(\dfrac{1 – \cos^2\theta}{\cos^2\theta}\right) = \sin^2\theta\cdot\dfrac{\sin^2\theta}{\cos^2\theta} = \sin^2\theta\tan^2\theta$ = RHS.

Notation and History

The English word “sine” arose from a chain of translation errors. The Indian mathematician Aryabhata (5th century CE) used jya (Sanskrit for “bowstring”), referring to the chord-half of a circle’s arc. Arabic translators rendered jya phonetically as jiba, which was later misread as jaib (meaning “fold” or “bay” in Arabic). Twelfth-century European translators rendered jaib into Latin as sinus (Latin for “fold” or “bay”) — and so the modern term “sine” is, etymologically, a 1500-year-old translation accident.

The other names have clearer origins. “Cosine” is short for complementi sinus — “sine of the complement.” “Tangent” comes from the Latin tangere (“to touch”) because the tangent line touches a circle at one point. “Secant” from secare (“to cut”). The cosecant and cotangent are simply the “co-functions” of cosecant and tangent — i.e. the corresponding ratio for the complementary angle.

The modern symbols $\sin, \cos, \tan$ were introduced by Leonhard Euler (Switzerland, 18th century). Indian mathematicians produced the first sine table around the 5th century; Arab and Persian astronomers refined it for navigation and timekeeping, and these tables travelled into medieval Europe through Spain. Without trigonometry, modern science — surveying, optics, mechanical waves, signal processing, GPS, computer graphics — would not exist.

Why Trigonometry Matters

Trigonometry began with ancient astronomers (Hipparchus, Aryabhata, Brahmagupta, Bhaskara) who needed to measure unreachable distances — the height of a mountain, the distance to the moon, the size of the Earth. The same six ratios are used today by surveyors marking property boundaries, civil engineers designing bridges, navigators on ships and aircraft, architects laying out a building plan, and physicists analysing waves and oscillations. The chapter you have just completed is the gateway to all of these applications. In Chapter 9 (Some Applications of Trigonometry) you will use these ratios to solve “heights and distances” problems — the practical face of trigonometry.

This concludes the solutions for ASSEB Class 10 Mathematics Chapter 8 — Introduction to Trigonometry. Practice each identity and the standard-angle values until you can write them down from memory; this fluency is essential for Chapter 9 (Some Applications of Trigonometry) and the heights-and-distances problems that appear in the HSLC examination. Keep visiting HSLC Guru for solutions to the rest of the Class 10 Mathematics syllabus.