Class 10 NCERT Solutions
Chapter 8: Introduction to Trigonometry
Master the trigonometric ratios, the standard angle values, and the fundamental identities of right-angled triangles with our step-by-step conceptual logic.
Exercise 8.1
1. In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine:
(i) $\sin A,\ \cos A$ (ii) $\sin C,\ \cos C$
First find the hypotenuse AC using the Pythagorean theorem:
$$\text{AC} = \sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25 \text{ cm}$$(i)
$$\sin A = \frac{\text{BC}}{\text{AC}} = \frac{7}{25}, \quad \cos A = \frac{\text{AB}}{\text{AC}} = \frac{24}{25}$$(ii)
$$\sin C = \frac{\text{AB}}{\text{AC}} = \frac{24}{25}, \quad \cos C = \frac{\text{BC}}{\text{AC}} = \frac{7}{25}$$
2. In figure, find $\tan P – \cot R$. (Given $PQ = 12$ cm, $PR = 13$ cm)
Find RQ using the Pythagorean theorem:
$$\text{RQ} = \sqrt{\text{PR}^2-\text{PQ}^2}=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5 \text{ cm}$$ $$\tan P = \frac{\text{RQ}}{\text{PQ}} = \frac{5}{12}, \quad \cot R = \frac{\text{RQ}}{\text{PQ}} = \frac{5}{12}$$Since $\tan P = \cot R$, we get $\tan P – \cot R = \mathbf{0}$.
3. If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
Let the sides be in the ratio $\sin A = \dfrac{3k}{4k}$. Find the adjacent side:
$$\text{AB} = \sqrt{(4k)^2-(3k)^2}=\sqrt{16k^2-9k^2}=\sqrt{7}\,k$$ $$\cos A = \frac{\text{AB}}{\text{AC}} = \frac{\sqrt{7}\,k}{4k} = \frac{\sqrt{7}}{4}$$ $$\tan A = \frac{\text{BC}}{\text{AB}} = \frac{3k}{\sqrt{7}\,k} = \frac{3}{\sqrt{7}}$$
4. Given $15\cot A = 8$, find $\sin A$ and $\sec A$.
From $15\cot A = 8$, we get $\cot A = \dfrac{8}{15}$. Let $\text{AB} = 8k$ and $\text{BC} = 15k$, then:
$$\text{AC} = \sqrt{(8k)^2+(15k)^2}=\sqrt{64k^2+225k^2}=\sqrt{289k^2}=17k$$ $$\sin A = \frac{\text{BC}}{\text{AC}} = \frac{15k}{17k} = \frac{15}{17}$$ $$\sec A = \frac{\text{AC}}{\text{AB}} = \frac{17k}{8k} = \frac{17}{8}$$
5. Given $\sec\theta = \frac{13}{12}$, calculate all other trigonometric ratios.
$\sec\theta = \dfrac{13}{12}$ means $\dfrac{\text{AC}}{\text{AB}} = \dfrac{13k}{12k}$. Find BC:
$$\text{BC} = \sqrt{(13k)^2-(12k)^2}=\sqrt{169k^2-144k^2}=5k$$ $$\sin\theta = \frac{5k}{13k} = \frac{5}{13}, \quad \cos\theta = \frac{12k}{13k} = \frac{12}{13}$$ $$\tan\theta = \frac{5k}{12k} = \frac{5}{12}, \quad \cot\theta = \frac{12k}{5k} = \frac{12}{5}$$ $$\operatorname{cosec}\theta = \frac{13k}{5k} = \frac{13}{5}$$
6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
Case I: Both angles belong to the same right-angled $\triangle$ABC.
$\cos A = \dfrac{\text{AC}}{\text{AB}}$ and $\cos B = \dfrac{\text{BC}}{\text{AB}}$. If $\cos A = \cos B$, then $\text{AC} = \text{BC}$, so $\angle B = \angle A$ (angles opposite equal sides are equal).
Case II: $\angle A$ belongs to $\triangle$APQ and $\angle B$ belongs to $\triangle$BXY.
If $\cos A = \cos B$, then $\dfrac{\text{AP}}{\text{AQ}} = \dfrac{\text{BX}}{\text{BY}}$, which gives $\dfrac{\text{AP}}{\text{BX}} = \dfrac{\text{AQ}}{\text{BY}} = k$ (say).
It follows that $\dfrac{\text{PQ}}{\text{XY}} = k$ as well, so $\triangle\text{APQ} \sim \triangle\text{BXY}$ by SSS similarity, giving $\angle A = \angle B$.
7. If $\cot\theta = \frac{7}{8}$, evaluate:
(i) $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
(ii) $\cot^2\theta$
(i) Simplify using the difference-of-squares identity:
$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$(ii)
$$\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$
8. If $3\cot A = 4$, check whether $\frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A – \sin^2 A$ or not.
$3\cot A = 4 \Rightarrow \cot A = \dfrac{4}{3}$, so $\tan A = \dfrac{3}{4}$. With $\text{AB} = 4k$, $\text{BC} = 3k$, $\text{AC} = 5k$:
$$\cos A = \frac{4}{5}, \quad \sin A = \frac{3}{5}$$ $$\text{LHS} = \frac{1-\frac{9}{16}}{1+\frac{9}{16}} = \frac{16-9}{16+9} = \frac{7}{25}$$ $$\text{RHS} = \left(\frac{4}{5}\right)^2 – \left(\frac{3}{5}\right)^2 = \frac{16-9}{25} = \frac{7}{25}$$Since LHS $=$ RHS, the identity is verified.
9. In triangle $ABC$, right-angled at $B$, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:
(i) $\sin A\cos C + \cos A\sin C$
(ii) $\cos A\cos C – \sin A\sin C$
From $\tan A = \dfrac{1}{\sqrt{3}}$, let $\text{AB} = \sqrt{3}\,k$, $\text{BC} = k$, then $\text{AC} = 2k$.
Note that $\sin A = \cos C = \dfrac{1}{2}$ and $\cos A = \sin C = \dfrac{\sqrt{3}}{2}$.
(i)
$$\sin A\cos C + \cos A\sin C = \frac{1}{2}\cdot\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{1}{4}+\frac{3}{4} = 1$$(ii)
$$\cos A\cos C – \sin A\sin C = \frac{\sqrt{3}}{2}\cdot\frac{1}{2} – \frac{1}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4} = 0$$
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 $\tan P$.
Let $\text{RQ} = x$, then $\text{PR} = 25 – x$. Applying the Pythagorean theorem:
$$(25-x)^2 = x^2 + 5^2 \Rightarrow 625-50x+x^2 = x^2+25$$ $$\Rightarrow 50x = 600 \Rightarrow x = 12$$So $\text{RQ} = 12$ cm and $\text{PR} = 13$ cm.
$$\sin P = \frac{12}{13}, \quad \cos P = \frac{5}{13}, \quad \tan P = \frac{12}{5}$$
11. State whether the following are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A = \frac{12}{5}$ 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) $\sin\theta = \frac{4}{3}$ for some angle $\theta$.
(i) False. $\tan A = \dfrac{\text{perpendicular}}{\text{base}}$, and the perpendicular is not always less than the base in a right triangle. For example, $\tan 60° = \sqrt{3} > 1$.
(ii) True. $\sec A = \dfrac{12}{5}$ is valid since $\cos A = \dfrac{5}{12} < 1$, which is perfectly acceptable.
(iii) False. $\cos A$ stands for the cosine of angle A, not the cosecant.
(iv) False. $\cot A$ is a single trigonometric ratio, not a product of “cot” and “A”.
(v) False. The sine of any angle always lies between $-1$ and $1$, so $\sin\theta = \dfrac{4}{3} > 1$ is impossible.
Exercise 8.2
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) $\frac{\cos 45°}{\sec 30° + \operatorname{cosec} 30°}$
(iv) $\frac{\sin 30° + \tan 45° – \operatorname{cosec} 60°}{\sec 30° + \cos 60° + \cot 45°}$
(v) $\frac{5\cos^2 60° + 4\sec^2 30° – \tan^2 45°}{\sin^2 30° + \cos^2 30°}$
Substitute the standard values and simplify each expression.
(i)
$$\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4}+\frac{1}{4} = 1$$(ii)
$$2(1)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2 = 2+\frac{3}{4}-\frac{3}{4} = 2$$(iii)
$$\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2} = \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)} \times \frac{(\sqrt{3}-1)\sqrt{2}}{(\sqrt{3}-1)\sqrt{2}} = \frac{(3-\sqrt{3})\sqrt{2}}{2\times2\times(3-1)} = \frac{3\sqrt{2}-\sqrt{6}}{8}$$(iv)
$$\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1} = \frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4} = \frac{(3\sqrt{3}-4)^2}{27-16} = \frac{27+16-24\sqrt{3}}{11} = \frac{43-24\sqrt{3}}{11}$$(v)
$$\frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\frac{5}{4}+\frac{16}{3}-1}{1} = \frac{15+64-12}{12} = \frac{67}{12}$$
2. Choose the correct option and justify your choice:
(i) $\frac{2\tan 30°}{1+\tan^2 30°} =$ (A) $\sin 60°$ (B) $\cos 60°$ (C) $\tan 60°$ (D) $\sin 30°$
(ii) $\frac{1-\tan^2 45°}{1+\tan^2 45°} =$ (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) $\frac{2\tan 30°}{1-\tan^2 30°} =$ (A) $\cos 60°$ (B) $\sin 60°$ (C) $\tan 60°$ (D) $\sin 30°$
(i) Answer: (A)
$$\frac{2\cdot\frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}}\times\frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60°$$(ii) Answer: (D)
$$\frac{1-(1)^2}{1+(1)^2} = \frac{0}{2} = 0$$(iii) Answer: (A) At $A = 0°$: $\sin 2(0°) = \sin 0° = 0$ and $2\sin 0° = 0$. Both sides are equal only when $A = 0°$.
(iv) Answer: (C)
$$\frac{2\times\frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}}\times\frac{3}{2} = \sqrt{3} = \tan 60°$$
3. If $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \frac{1}{\sqrt{3}}$; $0° < A+B \leq 90°$; $A > B$, find $A$ and $B$.
Match each expression to a known tangent value:
$$\tan(A+B) = \sqrt{3} = \tan 60° \Rightarrow A+B = 60° \quad \cdots(i)$$ $$\tan(A-B) = \frac{1}{\sqrt{3}} = \tan 30° \Rightarrow A-B = 30° \quad \cdots(ii)$$Adding (i) and (ii): $2A = 90° \Rightarrow A = 45°$. Substituting back: $B = 15°$.
4. State whether the following are true or false. Justify your answer:
(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 $A = 0°$.
(i) False. Counter-example with $A = 30°$, $B = 60°$:
$$\sin(30°+60°) = \sin 90° = 1, \quad \sin 30°+\sin 60° = \frac{1}{2}+\frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2} \neq 1$$(ii) True. $\sin 0° = 0$, $\sin 30° = \dfrac{1}{2}$, $\sin 90° = 1$ — the value increases as $\theta$ increases from $0°$ to $90°$.
(iii) False. $\cos 0° = 1$, $\cos 60° = \dfrac{1}{2}$, $\cos 90° = 0$ — the value decreases as $\theta$ increases.
(iv) False. $\sin\theta = \cos\theta$ only when $\theta = 45°$, not for all values.
(v) True. $\cot 0° = \dfrac{\cos 0°}{\sin 0°} = \dfrac{1}{0}$, which is undefined.
Exercise 8.3
1. Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
(i) Using $\operatorname{cosec}^2 A = 1 + \cot^2 A$:
$$\sin A = \frac{1}{\operatorname{cosec} A} = \frac{1}{\sqrt{1+\cot^2 A}}$$(ii) Using $\sec^2 A = 1 + \tan^2 A$ and $\tan A = \dfrac{1}{\cot A}$:
$$\sec A = \sqrt{1+\tan^2 A} = \sqrt{1+\frac{1}{\cot^2 A}} = \frac{\sqrt{\cot^2 A+1}}{\cot A}$$(iii)
$$\tan A = \frac{1}{\cot A}$$
2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
Starting from $\cos A = \dfrac{1}{\sec A}$ and using the Pythagorean identities:
$$\sin A = \frac{\sqrt{\sec^2 A-1}}{\sec A}, \quad \cos A = \frac{1}{\sec A}$$ $$\tan A = \sqrt{\sec^2 A-1}, \quad \cot A = \frac{1}{\sqrt{\sec^2 A-1}}$$ $$\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A-1}}$$
3. Evaluate:
(i) $\frac{\sin^2 63° + \sin^2 27°}{\cos^2 17° + \cos^2 73°}$
(ii) $\sin 25°\cos 65° + \cos 25°\sin 65°$
(i) Use the complementary angle relationships $\sin(90°-\theta) = \cos\theta$:
$$\frac{\sin^2(90°-27°)+\sin^2 27°}{\cos^2 17°+\cos^2(90°-17°)} = \frac{\cos^2 27°+\sin^2 27°}{\cos^2 17°+\sin^2 17°} = \frac{1}{1} = 1 \quad \left[\because \sin^2\theta+\cos^2\theta=1\right]$$(ii) Rewrite using complementary angles:
$$\sin 25°\cos(90°-25°)+\cos 25°\sin(90°-25°) = \sin^2 25°+\cos^2 25° = 1$$
4. Choose the correct option. Justify your choice.
(i) $9\sec^2 A – 9\tan^2 A =$ (A) 1 (B) 9 (C) 8 (D) 0
(ii) $(1+\tan\theta+\sec\theta)(1+\cot\theta-\operatorname{cosec}\theta) =$ (A) 0 (B) 1 (C) 2 (D) $-1$
(iii) $(\sec A+\tan A)(1-\sin A) =$ (A) $\sec A$ (B) $\sin A$ (C) $\operatorname{cosec} A$ (D) $\cos A$
(iv) $\frac{1+\tan^2 A}{1+\cot^2 A} =$ (A) $\sec^2 A$ (B) $-1$ (C) $\cot^2 A$ (D) $\tan^2 A$
(i) Answer: (B)
$$9\sec^2 A – 9\tan^2 A = 9(\sec^2 A – \tan^2 A) = 9 \times 1 = 9$$(ii) Answer: (C)
$$\left(\frac{\cos\theta+\sin\theta+1}{\cos\theta}\right)\left(\frac{\sin\theta+\cos\theta-1}{\sin\theta}\right) = \frac{(\cos\theta+\sin\theta)^2-1}{\cos\theta\sin\theta} = \frac{1+2\sin\theta\cos\theta-1}{\cos\theta\sin\theta} = 2$$(iii) Answer: (D)
$$\frac{1+\sin A}{\cos A}\cdot(1-\sin A) = \frac{1-\sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$$(iv) Answer: (D)
$$\frac{1+\tan^2 A}{1+\cot^2 A} = \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
5. Prove the following identities:
(i) $(\operatorname{cosec}\theta – \cot\theta)^2 = \frac{1-\cos\theta}{1+\cos\theta}$
(ii) $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A} = 2\sec A$
(iii) $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta} = 1+\sec\theta\operatorname{cosec}\theta$
(iv) $\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
(v) $\frac{\cos A-\sin A+1}{\cos A+\sin A-1} = \operatorname{cosec} A+\cot A$
(vi) $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A+\tan A$
(vii) $\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta} = \tan\theta$
(viii) $(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2 = 7+\tan^2 A+\cot^2 A$
(ix) $(\operatorname{cosec} A-\sin A)(\sec A-\cos A) = \frac{1}{\tan A+\cot A}$
(x) $\frac{1+\tan^2 A}{1+\cot^2 A} = \left(\frac{1-\tan A}{1-\cot A}\right)^2 = \tan^2 A$
(i)
$$\text{LHS} = \left(\frac{1-\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{1-\cos^2\theta} = \frac{(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{1-\cos\theta}{1+\cos\theta} = \text{RHS}$$(ii)
$$\text{LHS} = \frac{\cos^2 A+(1+\sin A)^2}{(1+\sin A)\cos A} = \frac{\cos^2 A+1+\sin^2 A+2\sin A}{(1+\sin A)\cos A} = \frac{2(1+\sin A)}{(1+\sin A)\cos A} = \frac{2}{\cos A} = 2\sec A = \text{RHS}$$(iii)
$$\text{LHS} = \frac{\sin^2\theta}{(\sin\theta-\cos\theta)\cos\theta} + \frac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)} = \frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}$$ $$= \frac{(\sin\theta-\cos\theta)(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta)}{\sin\theta\cos\theta(\sin\theta-\cos\theta)} = \frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta} = \operatorname{cosec}\theta\sec\theta+1 = \text{RHS}$$(iv)
$$\text{RHS} = \frac{\sin^2 A}{1-\cos A} = \frac{1-\cos^2 A}{1-\cos A} = \frac{(1+\cos A)(1-\cos A)}{1-\cos A} = 1+\cos A = 1+\frac{1}{\sec A} = \frac{\sec A+1}{\sec A} = \text{LHS}$$(v) Divide numerator and denominator by $\sin A$:
$$\text{LHS} = \frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1} = \frac{(\cot A+\operatorname{cosec} A)(1-(\operatorname{cosec} A-\cot A))}{\cot A-\operatorname{cosec} A+1} = \cot A+\operatorname{cosec} A = \text{RHS}$$(vi)
$$\text{LHS} = \sqrt{\frac{(1+\sin A)^2}{(1-\sin A)(1+\sin A)}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} = \frac{1+\sin A}{\cos A} = \sec A+\tan A = \text{RHS}$$(vii)
$$\text{LHS} = \frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)} = \frac{\sin\theta(\cos^2\theta-\sin^2\theta)}{\cos\theta(\cos^2\theta-\sin^2\theta)} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$$(viii)
$$\text{LHS} = \sin^2 A+\operatorname{cosec}^2 A+2+\cos^2 A+\sec^2 A+2 = 1+(1+\cot^2 A)+2+(1+\tan^2 A)+2 = 7+\tan^2 A+\cot^2 A = \text{RHS}$$(ix)
$$\text{LHS} = \frac{1-\sin^2 A}{\sin A}\cdot\frac{1-\cos^2 A}{\cos A} = \frac{\cos^2 A\cdot\sin^2 A}{\sin A\cos A} = \sin A\cos A = \frac{\sin A\cos A}{\sin^2 A+\cos^2 A} = \frac{1}{\tan A+\cot A} = \text{RHS}$$(x)
$$\frac{1+\tan^2 A}{1+\cot^2 A} = \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A = \text{RHS}$$ $$\left(\frac{1-\tan A}{1-\cot A}\right)^2 = \left(\frac{\tan A(1-\tan A)}{\tan A-1}\right)^2 = (-\tan A)^2 = \tan^2 A = \text{RHS}$$