NCERT Solutions Class 10 Maths Chapter 8 Ex 8.4 | Introduction to Trigonometry NCERT Class 10 Maths – Exercise 8.3 Solutions

NCERT Class 10 Maths

Chapter 8 – Introduction to Trigonometry | Exercise 8.3

(Rationalized Syllabus 2025-26)

💡 Fundamental Identities

  • $\sin^2 A + \cos^2 A = 1$
  • $1 + \tan^2 A = \sec^2 A$
  • $1 + \cot^2 A = \text{cosec}^2 A$
Q1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

(i) tan A
By definition:
$$\tan A = \frac{1}{\cot A}$$
(ii) sec A
Using identity $1 + \tan^2 A = \sec^2 A$:
$$\sec^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$$
$$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$$
(iii) sin A
Using identity $1 + \cot^2 A = \text{cosec}^2 A$:
$$\text{cosec}^2 A = 1 + \cot^2 A \Rightarrow \text{cosec } A = \sqrt{1 + \cot^2 A}$$
$$\sin A = \frac{1}{\text{cosec } A} = \frac{1}{\sqrt{1 + \cot^2 A}}$$
Q2

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

1. $\cos A = \frac{1}{\sec A}$
2. $\sin^2 A = 1 – \cos^2 A = 1 – \frac{1}{\sec^2 A} = \frac{\sec^2 A – 1}{\sec^2 A} \Rightarrow \sin A = \frac{\sqrt{\sec^2 A – 1}}{\sec A}$
3. $\tan^2 A = \sec^2 A – 1 \Rightarrow \tan A = \sqrt{\sec^2 A – 1}$
4. $\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A – 1}}$
5. $\text{cosec } A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A – 1}}$
Q3

Choose the correct option. Justify your choice.

(i) $9 \sec^2 A – 9 \tan^2 A = $
$$= 9(\sec^2 A – \tan^2 A)$$
Since $\sec^2 A – \tan^2 A = 1$:
$$= 9(1) = 9$$
✔️ (B) 9
(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \text{cosec } \theta) = $
Convert to sin and cos:
$$= (1 + \frac{S}{C} + \frac{1}{C})(1 + \frac{C}{S} – \frac{1}{S}) = (\frac{C+S+1}{C})(\frac{S+C-1}{S})$$
$$= \frac{((S+C)+1)((S+C)-1)}{SC} = \frac{(S+C)^2 – 1^2}{SC}$$
$$= \frac{S^2 + C^2 + 2SC – 1}{SC} = \frac{1 + 2SC – 1}{SC} = \frac{2SC}{SC} = 2$$
✔️ (C) 2
(iii) $(\sec A + \tan A)(1 – \sin A) = $
$$= (\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 – \sin A) = \frac{1 + \sin A}{\cos A} \times (1 – \sin A)$$
$$= \frac{1 – \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$$
✔️ (D) cos A
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} = $
$$= \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
✔️ (D) $\tan^2 A$
Q4

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

(i) $(\text{cosec } \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta}$
LHS:
$$= \left(\frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1 – \cos \theta}{\sin \theta}\right)^2 = \frac{(1 – \cos \theta)^2}{\sin^2 \theta}$$
$$= \frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta} = \frac{(1 – \cos \theta)(1 – \cos \theta)}{(1 – \cos \theta)(1 + \cos \theta)} = \frac{1 – \cos \theta}{1 + \cos \theta} = \text{RHS}$$
✔️ Proved
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
LHS:
$$= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)} = \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{\cos A(1 + \sin A)}$$
$$= \frac{1 + 1 + 2\sin A}{\cos A(1 + \sin A)} = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A = \text{RHS}$$
✔️ Proved
(iii) $\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \text{cosec } \theta$
LHS: Convert to sin/cos.
$$= \frac{\frac{\sin}{\cos}}{1 – \frac{\cos}{\sin}} + \frac{\frac{\cos}{\sin}}{1 – \frac{\sin}{\cos}} = \frac{\sin^2}{\cos(\sin – \cos)} + \frac{\cos^2}{\sin(\cos – \sin)}$$
$$= \frac{\sin^2}{\cos(\sin – \cos)} – \frac{\cos^2}{\sin(\sin – \cos)} = \frac{\sin^3 – \cos^3}{\sin \cos (\sin – \cos)}$$
Use $a^3 – b^3 = (a-b)(a^2 + ab + b^2)$:
$$= \frac{(\sin – \cos)(\sin^2 + \cos^2 + \sin \cos)}{\sin \cos (\sin – \cos)} = \frac{1 + \sin \cos}{\sin \cos}$$
$$= \frac{1}{\sin \cos} + 1 = \text{cosec } \theta \sec \theta + 1 = \text{RHS}$$
✔️ Proved
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$
LHS: $\frac{1 + 1/\cos}{1/\cos} = \frac{(\cos+1)/\cos}{1/\cos} = \cos A + 1$.
RHS: $\frac{1 – \cos^2 A}{1 – \cos A} = \frac{(1-\cos A)(1+\cos A)}{1-\cos A} = 1 + \cos A$.
✔️ LHS = RHS
(v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$
Divide Numerator and Denominator by $\sin A$:
$$= \frac{\cot A – 1 + \text{cosec } A}{\cot A + 1 – \text{cosec } A} = \frac{(\cot A + \text{cosec } A) – 1}{1 – \text{cosec } A + \cot A}$$
Replace $1$ with $\text{cosec}^2 A – \cot^2 A$:
$$= \frac{(\cot + \text{cosec}) – (\text{cosec}^2 – \cot^2)}{1 – \text{cosec} + \cot}$$
$$= \frac{(\cot + \text{cosec})[1 – (\text{cosec} – \cot)]}{1 – \text{cosec} + \cot} = \cot A + \text{cosec } A$$
✔️ Proved
(vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
Rationalize inside root:
$$= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 – \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$$
$$= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$$
✔️ Proved
(vii) $\frac{\sin \theta – 2 \sin^3 \theta}{2 \cos^3 \theta – \cos \theta} = \tan \theta$
LHS:
$$= \frac{\sin \theta(1 – 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta – 1)}$$
Using $\cos 2\theta$ forms (or simplify): $1 – 2\sin^2 \theta = \cos^2 – \sin^2$ and $2\cos^2 – 1 = \cos^2 – \sin^2$.
So the bracket terms cancel out.
$$= \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS}$$
✔️ Proved
(viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
Expand:
$$= (\sin^2 + \text{cosec}^2 + 2) + (\cos^2 + \sec^2 + 2)$$
$$= (\sin^2 + \cos^2) + \text{cosec}^2 + \sec^2 + 4$$
$$= 1 + (1 + \cot^2) + (1 + \tan^2) + 4$$
$$= 1 + 1 + 1 + 4 + \cot^2 + \tan^2 = 7 + \tan^2 A + \cot^2 A$$
✔️ Proved
(ix) $(\text{cosec } A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$
LHS: $(\frac{1}{\sin} – \sin)(\frac{1}{\cos} – \cos) = (\frac{1-\sin^2}{\sin})(\frac{1-\cos^2}{\cos}) = \frac{\cos^2 \sin^2}{\sin \cos} = \sin A \cos A$.
RHS: $\frac{1}{\frac{\sin}{\cos} + \frac{\cos}{\sin}} = \frac{1}{\frac{\sin^2 + \cos^2}{\sin \cos}} = \frac{\sin \cos}{1} = \sin A \cos A$.
✔️ LHS = RHS
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$
Part 1: $\frac{\sec^2 A}{\text{cosec}^2 A} = \tan^2 A$ (See Q3 iv).
Part 2: $\left(\frac{1 – \tan A}{1 – 1/\tan A}\right)^2 = \left(\frac{1 – \tan A}{\frac{\tan A – 1}{\tan A}}\right)^2 = \left(\frac{(1-\tan A)\tan A}{-(1-\tan A)}\right)^2$
$$= (-\tan A)^2 = \tan^2 A$$
✔️ Proved
🎉 Exercise 8.3 Completed | Chapter 8 Finished!
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