NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3

Q1: Prove $\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} – \tan^2\frac{\pi}{4} = -\frac{1}{2}$

Values: $\sin\frac{\pi}{6} = \frac{1}{2}, \cos\frac{\pi}{3} = \frac{1}{2}, \tan\frac{\pi}{4} = 1$

LHS $= (\frac{1}{2})^2 + (\frac{1}{2})^2 – (1)^2$

$= \frac{1}{4} + \frac{1}{4} – 1 = \frac{2}{4} – 1 = \frac{1}{2} – 1$

$= -\frac{1}{2} = \text{RHS}$

Q2: Prove $2\sin^2\frac{\pi}{6} + \text{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3} = \frac{3}{2}$

Values: $\sin\frac{\pi}{6} = \frac{1}{2}, \cos\frac{\pi}{3} = \frac{1}{2}$

$\text{cosec}\frac{7\pi}{6} = \text{cosec}(\pi + \frac{\pi}{6}) = -\text{cosec}\frac{\pi}{6} = -2$

LHS $= 2(\frac{1}{2})^2 + (-2)^2(\frac{1}{2})^2$

$= 2(\frac{1}{4}) + 4(\frac{1}{4}) = \frac{1}{2} + 1$

$= \frac{3}{2} = \text{RHS}$

Q3: Prove $\cot^2\frac{\pi}{6} + \text{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6$

$\cot\frac{\pi}{6} = \sqrt{3}, \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$

$\text{cosec}\frac{5\pi}{6} = \text{cosec}(\pi – \frac{\pi}{6}) = \text{cosec}\frac{\pi}{6} = 2$

LHS $= (\sqrt{3})^2 + 2 + 3(\frac{1}{\sqrt{3}})^2$

$= 3 + 2 + 3(\frac{1}{3}) = 3 + 2 + 1$

$= 6 = \text{RHS}$

Q4: Prove $2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10$

$\sin\frac{3\pi}{4} = \sin(\pi – \frac{\pi}{4}) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \sec\frac{\pi}{3} = 2$

LHS $= 2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2$

$= 2(\frac{1}{2}) + 2(\frac{1}{2}) + 2(4) = 1 + 1 + 8$

$= 10 = \text{RHS}$

Q5: Find value of (i) sin 75° (ii) tan 15°

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

(i) $\sin 75^\circ$: $\sin(45^\circ + 30^\circ)$
$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

$\tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$

(ii) $\tan 15^\circ$: $\tan(45^\circ – 30^\circ)$
$= \frac{\tan 45^\circ – \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$
$= \frac{\sqrt{3}-1}{\sqrt{3}+1}$. Rationalize: $\frac{(\sqrt{3}-1)^2}{2} = \frac{3+1-2\sqrt{3}}{2} = 2-\sqrt{3}$

Answers: (i) $\frac{\sqrt{3}+1}{2\sqrt{2}}$ (ii) $2-\sqrt{3}$

Q6: Prove $\cos(\frac{\pi}{4}-x)\cos(\frac{\pi}{4}-y) – \sin(\frac{\pi}{4}-x)\sin(\frac{\pi}{4}-y) = \sin(x+y)$

$\cos A \cos B – \sin A \sin B = \cos(A+B)$

Let $A = \frac{\pi}{4}-x$ and $B = \frac{\pi}{4}-y$.

LHS $= \cos[(\frac{\pi}{4}-x) + (\frac{\pi}{4}-y)]$

$= \cos[\frac{\pi}{2} – (x+y)]$

$= \sin(x+y) = \text{RHS}$

Q7: Prove $\frac{\tan(\frac{\pi}{4}+x)}{\tan(\frac{\pi}{4}-x)} = \left(\frac{1+\tan x}{1-\tan x}\right)^2$

$\tan(\frac{\pi}{4}+A) = \frac{1+\tan A}{1-\tan A}, \tan(\frac{\pi}{4}-A) = \frac{1-\tan A}{1+\tan A}$

LHS $= \frac{\left(\frac{1+\tan x}{1-\tan x}\right)}{\left(\frac{1-\tan x}{1+\tan x}\right)}$

$= \frac{1+\tan x}{1-\tan x} \times \frac{1+\tan x}{1-\tan x}$

$= \left(\frac{1+\tan x}{1-\tan x}\right)^2 = \text{RHS}$

Q8: Prove $\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)} = \cot^2 x$

Reductions:
$\cos(\pi+x) = -\cos x$, $\cos(-x) = \cos x$
$\sin(\pi-x) = \sin x$, $\cos(\frac{\pi}{2}+x) = -\sin x$

LHS $= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$

$= \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x}$

$= \cot^2 x = \text{RHS}$

Q9: Prove $\cos(\frac{3\pi}{2}+x)\cos(2\pi+x)[\cot(\frac{3\pi}{2}-x) + \cot(2\pi+x)] = 1$

Reductions:
$\cos(\frac{3\pi}{2}+x) = \sin x$, $\cos(2\pi+x) = \cos x$
$\cot(\frac{3\pi}{2}-x) = \tan x$, $\cot(2\pi+x) = \cot x$

LHS $= \sin x \cos x [\tan x + \cot x]$

$= \sin x \cos x [\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}]$
$= \sin x \cos x [\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}]$

$= \sin^2 x + \cos^2 x = 1 = \text{RHS}$

Q10: Prove $\sin(n+1)x\sin(n+2)x + \cos(n+1)x\cos(n+2)x = \cos x$

$\cos A \cos B + \sin A \sin B = \cos(A-B)$

Let $A = (n+2)x$ and $B = (n+1)x$.

LHS $= \cos[(n+2)x – (n+1)x]$

$= \cos[nx + 2x – nx – x]$

$= \cos x = \text{RHS}$

Q11: Prove $\cos(\frac{3\pi}{4}+x) – \cos(\frac{3\pi}{4}-x) = -\sqrt{2}\sin x$

$\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$

LHS $= -2\sin\left(\frac{\frac{3\pi}{4}+x+\frac{3\pi}{4}-x}{2}\right) \sin\left(\frac{\frac{3\pi}{4}+x-(\frac{3\pi}{4}-x)}{2}\right)$

$= -2\sin(\frac{3\pi}{4}) \sin(x)$

$\sin\frac{3\pi}{4} = \sin(\pi-\frac{\pi}{4}) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$= -2(\frac{1}{\sqrt{2}})\sin x = -\sqrt{2}\sin x = \text{RHS}$

Q12: Prove $\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x$

$\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$

LHS $= \sin(6x+4x)\sin(6x-4x)$

$= \sin 10x \sin 2x = \text{RHS}$

Q13: Prove $\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$

$\cos^2 A – \cos^2 B = \sin(B+A)\sin(B-A)$

LHS $= \sin(6x+2x)\sin(6x-2x)$

$= \sin 8x \sin 4x = \text{RHS}$

Q14: Prove $\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x$

Group terms: $(\sin 6x + \sin 2x) + 2\sin 4x$

$= 2\sin 4x \cos 2x + 2\sin 4x$ (Using $\sin C + \sin D$)

$= 2\sin 4x (\cos 2x + 1)$

$1 + \cos 2x = 2\cos^2 x$

$= 2\sin 4x (2\cos^2 x) = 4\cos^2 x \sin 4x = \text{RHS}$

Q15: Prove $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x)$

LHS $= \frac{\cos 4x}{\sin 4x} (2\sin 4x \cos x) = 2\cos 4x \cos x$

RHS $= \frac{\cos x}{\sin x} (2\cos 4x \sin x) = 2\cos 4x \cos x$

LHS = RHS

Q16: Prove $\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

Num: $-2\sin \frac{14x}{2} \sin \frac{4x}{2} = -2\sin 7x \sin 2x$

Den: $2\cos \frac{20x}{2} \sin \frac{14x}{2} = 2\cos 10x \sin 7x$

Ratio: $\frac{-2\sin 7x \sin 2x}{2\cos 10x \sin 7x}$

$= -\frac{\sin 2x}{\cos 10x} = \text{RHS}$

Q17: Prove $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Num: $2\sin 4x \cos x$

Den: $2\cos 4x \cos x$

$= \frac{2\sin 4x \cos x}{2\cos 4x \cos x} = \tan 4x = \text{RHS}$

Q18: Prove $\frac{\sin x – \sin y}{\cos x + \cos y} = \tan \frac{x-y}{2}$

Num: $2\cos \frac{x+y}{2} \sin \frac{x-y}{2}$

Den: $2\cos \frac{x+y}{2} \cos \frac{x-y}{2}$

$= \frac{\sin \frac{x-y}{2}}{\cos \frac{x-y}{2}} = \tan \frac{x-y}{2} = \text{RHS}$

Q19: Prove $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$

Num: $2\sin 2x \cos (-x) = 2\sin 2x \cos x$

Den: $2\cos 2x \cos (-x) = 2\cos 2x \cos x$

$= \frac{\sin 2x}{\cos 2x} = \tan 2x = \text{RHS}$

Q20: Prove $\frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x} = 2\sin x$

Num: $2\cos 2x \sin(-x) = -2\cos 2x \sin x$

Den: $-(\cos^2 x – \sin^2 x) = -\cos 2x$

Ratio: $\frac{-2\cos 2x \sin x}{-\cos 2x}$

$= 2\sin x = \text{RHS}$

Q21: Prove $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Group $(\cos 4x + \cos 2x)$ and $(\sin 4x + \sin 2x)$.

Num: $2\cos 3x \cos x + \cos 3x = \cos 3x(2\cos x + 1)$

Den: $2\sin 3x \cos x + \sin 3x = \sin 3x(2\cos x + 1)$

$= \frac{\cos 3x}{\sin 3x} = \cot 3x = \text{RHS}$

Q22: Prove $\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$

$\cot 3x = \cot(2x+x) = \frac{\cot 2x \cot x – 1}{\cot 2x + \cot x}$

$\cot 3x (\cot 2x + \cot x) = \cot 2x \cot x – 1$

$\cot 3x \cot 2x + \cot 3x \cot x = \cot 2x \cot x – 1$

Rearranging gives the required identity. RHS = 1.

Q23: Prove $\tan 4x = \frac{4\tan x(1-\tan^2 x)}{1-6\tan^2 x + \tan^4 x}$

Let $A=2x$. $\tan 4x = \frac{2\tan 2x}{1-\tan^2 2x}$.

Substitute $\tan 2x = \frac{2\tan x}{1-\tan^2 x}$.

Num: $2(\frac{2\tan x}{1-\tan^2 x}) = \frac{4\tan x}{1-\tan^2 x}$.

Den: $1 – (\frac{2\tan x}{1-\tan^2 x})^2 = \frac{(1-\tan^2 x)^2 – 4\tan^2 x}{(1-\tan^2 x)^2}$.

Simplifying gives the RHS.

Q24: Prove $\cos 4x = 1 – 8\sin^2 x \cos^2 x$

$\cos 4x = 1 – 2\sin^2 2x$

$= 1 – 2(2\sin x \cos x)^2$

$= 1 – 2(4\sin^2 x \cos^2 x)$

$= 1 – 8\sin^2 x \cos^2 x = \text{RHS}$

Q25: Prove $\cos 6x = 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1$

$\cos 6x = \cos 3(2x) = 4\cos^3(2x) – 3\cos(2x)$

Substitute $\cos 2x = 2\cos^2 x – 1$.

$= 4(2\cos^2 x – 1)^3 – 3(2\cos^2 x – 1)$

Expand $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$.

$= 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1 = \text{RHS}$

Trigonometric Functions

1. Evaluation of Values (Q1-Q5)

2. Compound Angle Identities (Q6-Q10)

3. Sum to Product Identities (Q11-Q21)

4. Multiple Angle Identities (Q22-Q25)