Last Updated on November 24, 2025 by Aman Singh
Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 3 Exercise 3.3 . Master essential Trigonometric Identities including: Sum and Difference formulas (Q.5, Q.6, Q.7), Product-to-Sum formulas (Q.11, Q.12, Q.13), and Double/Triple Angle expansions (Q.24, Q.25). Practice proving identities using standard values, angle transformation, and algebraic manipulation. Essential for developing proof skills in trigonometry.
This exercise involves proving various trigonometric identities and finding exact values using sum, difference, double angle, and product-to-sum formulas.
Proving Identities Using Standard Values (Exercises 1-4)
1. Prove: $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} – \tan^2 \frac{\pi}{4} = -\frac{1}{2}$
We use the standard values: $\sin \frac{\pi}{6} = \frac{1}{2}$, $\cos \frac{\pi}{3} = \frac{1}{2}$, $\tan \frac{\pi}{4} = 1$.
$$\text{LHS} = \left(\sin \frac{\pi}{6}\right)^2 + \left(\cos \frac{\pi}{3}\right)^2 – \left(\tan \frac{\pi}{4}\right)^2$$
$$\text{LHS} = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 – (1)^2$$
$$\text{LHS} = \frac{1}{4} + \frac{1}{4} – 1$$
$$\text{LHS} = \frac{2}{4} – 1 = \frac{1}{2} – 1 = \mathbf{-\frac{1}{2}}$$
$$\text{LHS} = \text{RHS}$$
(Note: The provided RHS in the question is 2, but the calculation yields $-1/2$. Assuming the book intended to prove the expression equals $-1/2$.)
2. Prove: $2\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}$
We use the standard values and quadrant rules.
$$\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}$$
For $\csc \frac{7\pi}{6}$:
$$\csc \frac{7\pi}{6} = \csc \left(\pi + \frac{\pi}{6}\right)$$
In the third quadrant, $\csc$ is negative: $\csc (\pi + \theta) = -\csc \theta$.
$$\csc \left(\pi + \frac{\pi}{6}\right) = -\csc \frac{\pi}{6} = -2$$
Substitute values into LHS:
$$\text{LHS} = 2\left(\sin \frac{\pi}{6}\right)^2 + \left(\csc \frac{7\pi}{6}\right)^2 \left(\cos \frac{\pi}{3}\right)^2$$
$$\text{LHS} = 2\left(\frac{1}{2}\right)^2 + (-2)^2 \left(\frac{1}{2}\right)^2$$
$$\text{LHS} = 2\left(\frac{1}{4}\right) + 4\left(\frac{1}{4}\right)$$
$$\text{LHS} = \frac{1}{2} + 1 = \frac{3}{2}$$
$$\text{LHS} = \text{RHS}$$
3. Prove: $\cot^2 \frac{\pi}{6} + \csc \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$
We use the standard values and quadrant rules.
$$\cot \frac{\pi}{6} = \sqrt{3}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$
For $\csc \frac{5\pi}{6}$:
$$\csc \frac{5\pi}{6} = \csc \left(\pi – \frac{\pi}{6}\right)$$
In the second quadrant, $\csc$ is positive: $\csc (\pi – \theta) = \csc \theta$.
$$\csc \left(\pi – \frac{\pi}{6}\right) = \csc \frac{\pi}{6} = 2$$
Substitute values into LHS:
$$\text{LHS} = \left(\cot \frac{\pi}{6}\right)^2 + \csc \frac{5\pi}{6} + 3\left(\tan \frac{\pi}{6}\right)^2$$
$$\text{LHS} = (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2$$
$$\text{LHS} = 3 + 2 + 3\left(\frac{1}{3}\right)$$
$$\text{LHS} = 3 + 2 + 1 = \mathbf{6}$$
$$\text{LHS} = \text{RHS}$$
4. Prove: $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$
We use the standard values and quadrant rules.
$$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sec \frac{\pi}{3} = 2$$
For $\sin \frac{3\pi}{4}$:
$$\sin \frac{3\pi}{4} = \sin \left(\pi – \frac{\pi}{4}\right)$$
In the second quadrant, $\sin$ is positive: $\sin (\pi – \theta) = \sin \theta$.
$$\sin \left(\pi – \frac{\pi}{4}\right) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$$
Substitute values into LHS:
$$\text{LHS} = 2\left(\sin \frac{3\pi}{4}\right)^2 + 2\left(\cos \frac{\pi}{4}\right)^2 + 2\left(\sec \frac{\pi}{3}\right)^2$$
$$\text{LHS} = 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2(2)^2$$
$$\text{LHS} = 2\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) + 2(4)$$
$$\text{LHS} = 1 + 1 + 8 = \mathbf{10}$$
$$\text{LHS} = \text{RHS}$$
5. Find the value of:
We use the Sum/Difference Identities:
- $\sin(A+B) = \sin A \cos B + \cos A \sin B$
- $\tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$
(i) $\sin 75^{\circ}$
We can write $75^{\circ}$ as $45^{\circ} + 30^{\circ}$.
$$\sin 75^{\circ} = \sin (45^{\circ} + 30^{\circ})$$
$$\sin 75^{\circ} = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}$$
$$\sin 75^{\circ} = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$$
$$\sin 75^{\circ} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \mathbf{\frac{\sqrt{3} + 1}{2\sqrt{2}}}$$
(ii) $\tan 15^{\circ}$
We can write $15^{\circ}$ as $45^{\circ} – 30^{\circ}$ (or $60^{\circ} – 45^{\circ}$).
$$\tan 15^{\circ} = \tan (45^{\circ} – 30^{\circ})$$
$$\tan 15^{\circ} = \frac{\tan 45^{\circ} – \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}}$$
$$\tan 15^{\circ} = \frac{1 – 1/\sqrt{3}}{1 + (1)(1/\sqrt{3})} = \frac{(\sqrt{3} – 1)/\sqrt{3}}{(\sqrt{3} + 1)/\sqrt{3}}$$
$$\tan 15^{\circ} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}$$
Rationalize the denominator:
$$\tan 15^{\circ} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} – 1}{\sqrt{3} – 1} = \frac{(\sqrt{3} – 1)^2}{3 – 1}$$
$$\tan 15^{\circ} = \frac{3 – 2\sqrt{3} + 1}{2} = \frac{4 – 2\sqrt{3}}{2} = \mathbf{2 – \sqrt{3}}$$
Proving Identities (Exercises 6-25)
6. Prove: $\cos\left(\frac{\pi}{4} – x\right) \cos\left(\frac{\pi}{4} – y\right) – \sin\left(\frac{\pi}{4} – x\right) \sin\left(\frac{\pi}{4} – y\right) = \sin(x+y)$
This identity has the form $\cos A \cos B – \sin A \sin B$, which is the expansion of $\cos(A+B)$.
Let $A = \frac{\pi}{4} – x$ and $B = \frac{\pi}{4} – y$.
$$\text{LHS} = \cos(A+B)$$
$$\text{LHS} = \cos\left(\left(\frac{\pi}{4} – x\right) + \left(\frac{\pi}{4} – y\right)\right)$$
$$\text{LHS} = \cos\left(\frac{2\pi}{4} – (x + y)\right)$$
$$\text{LHS} = \cos\left(\frac{\pi}{2} – (x + y)\right)$$
Using the identity $\cos(\frac{\pi}{2} – \theta) = \sin \theta$:
$$\text{LHS} = \sin(x + y) = \text{RHS}$$
7. Prove: $\frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} – x)} = \left(\frac{1 + \tan x}{1 – \tan x}\right)^2$
We use the sum and difference formulas for tangent:
$$\tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}, \quad \tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$$
Let $A = \frac{\pi}{4}$, so $\tan A = 1$.
$$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 – \tan x}$$
$$\tan\left(\frac{\pi}{4} – x\right) = \frac{1 – \tan x}{1 + \tan x}$$
$$\text{LHS} = \frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} – x)} = \frac{(1 + \tan x) / (1 – \tan x)}{(1 – \tan x) / (1 + \tan x)}$$
$$\text{LHS} = \frac{1 + \tan x}{1 – \tan x} \times \frac{1 + \tan x}{1 – \tan x} = \left(\frac{1 + \tan x}{1 – \tan x}\right)^2 = \text{RHS}$$
8. Prove: $\frac{\cos(\pi + x) \cos(-x)}{\sin(\pi – x) \cos(\frac{\pi}{2} + x)} = \cot^2 x$
We use the angle transformation formulas:
- $\cos(\pi + x) = -\cos x$ (3rd quadrant)
- $\cos(-x) = \cos x$ (Even function)
- $\sin(\pi – x) = \sin x$ (2nd quadrant)
- $\cos(\frac{\pi}{2} + x) = -\sin x$ (2nd quadrant, $\cos \to \sin$)
Substitute these into the LHS:
$$\text{LHS} = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$$
$$\text{LHS} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x}$$
$$\text{LHS} = \left(\frac{\cos x}{\sin x}\right)^2 = \cot^2 x = \text{RHS}$$
9. Prove: $\cos\left(\frac{3\pi}{2} + x\right) \cos(2\pi + x) \left[\cot\left(\frac{3\pi}{2} – x\right) + \cot(2\pi + x)\right] = 1$
We use the angle transformation and periodicity formulas:
- $\cos(\frac{3\pi}{2} + x) = \sin x$ (4th quadrant, $\cos \to \sin$)
- $\cos(2\pi + x) = \cos x$ (Periodicity)
- $\cot(\frac{3\pi}{2} – x) = \tan x$ (3rd quadrant, $\cot \to \tan$)
- $\cot(2\pi + x) = \cot x$ (Periodicity)
Substitute these into the LHS:
$$\text{LHS} = (\sin x)(\cos x) [\tan x + \cot x]$$
$$\text{LHS} = \sin x \cos x \left[\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right]$$
Find a common denominator inside the bracket:
$$\text{LHS} = \sin x \cos x \left[\frac{\sin^2 x + \cos^2 x}{\cos x \sin x}\right]$$
Since $\sin^2 x + \cos^2 x = 1$:
$$\text{LHS} = \sin x \cos x \left[\frac{1}{\cos x \sin x}\right] = \mathbf{1} = \text{RHS}$$
10. Prove: $\sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x$
This identity has the form $\cos A \cos B + \sin A \sin B$, which is the expansion of $\cos(A – B)$.
Let $A = (n + 2)x$ and $B = (n + 1)x$.
$$\text{LHS} = \cos(A – B)$$
$$\text{LHS} = \cos((n + 2)x – (n + 1)x)$$
$$\text{LHS} = \cos(nx + 2x – nx – x)$$
$$\text{LHS} = \cos(x) = \text{RHS}$$
11. Prove: $\cos\left(\frac{3\pi}{4} + x\right) – \cos\left(\frac{3\pi}{4} – x\right) = -\sqrt{2} \sin x$
We use the $\cos C – \cos D$ Product-to-Sum Formula:
$$\cos C – \cos D = -2 \sin \left(\frac{C + D}{2}\right) \sin \left(\frac{C – D}{2}\right)$$
Let $C = \frac{3\pi}{4} + x$ and $D = \frac{3\pi}{4} – x$.
- $C + D = \left(\frac{3\pi}{4} + x\right) + \left(\frac{3\pi}{4} – x\right) = \frac{6\pi}{4} = \frac{3\pi}{2}$
- $C – D = \left(\frac{3\pi}{4} + x\right) – \left(\frac{3\pi}{4} – x\right) = 2x$
$$\text{LHS} = -2 \sin \left(\frac{3\pi/2}{2}\right) \sin \left(\frac{2x}{2}\right)$$
$$\text{LHS} = -2 \sin \left(\frac{3\pi}{4}\right) \sin x$$
Now, evaluate $\sin \left(\frac{3\pi}{4}\right) = \sin(\pi – \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
$$\text{LHS} = -2 \left(\frac{1}{\sqrt{2}}\right) \sin x$$
Since $\frac{2}{\sqrt{2}} = \sqrt{2}$:
$$\text{LHS} = -\sqrt{2} \sin x = \text{RHS}$$
12. Prove: $\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x$
We use the identity $\sin^2 A – \sin^2 B = \sin(A+B) \sin(A-B)$.
Here $A = 6x$ and $B = 4x$.
$$\text{LHS} = \sin^2 6x – \sin^2 4x$$
$$\text{LHS} = \sin(6x + 4x) \sin(6x – 4x)$$
$$\text{LHS} = \sin(10x) \sin(2x) = \text{RHS}$$
13. Prove: $\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$
We use the identity $\cos^2 A – \cos^2 B = \sin(B+A) \sin(B-A)$ or convert to $\sin^2$:
Use $\cos^2 A = 1 – \sin^2 A$.
$$\text{LHS} = (1 – \sin^2 2x) – (1 – \sin^2 6x)$$
$$\text{LHS} = 1 – \sin^2 2x – 1 + \sin^2 6x$$
$$\text{LHS} = \sin^2 6x – \sin^2 2x$$
Now use $\sin^2 A – \sin^2 B = \sin(A+B) \sin(A-B)$:
$$\text{LHS} = \sin(6x + 2x) \sin(6x – 2x)$$
$$\text{LHS} = \sin(8x) \sin(4x) = \text{RHS}$$
14. Prove: $\sin x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$
(Note: The question is likely intended to be $\sin 2x + \sin 4x + \sin 6x$ or similar, as written it doesn’t appear to simplify cleanly to the RHS. Assuming the question intended to be $\sin 2x + \sin 4x + \sin 6x$ or that there is a typo in the question and should be $2\sin 2x$ instead of $\sin x$. Let’s solve the slightly corrected identity $\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$ for completeness).
Assuming the question is $\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$:
$$\text{LHS} = (\sin 6x + \sin 2x) + 2 \sin 4x$$
Use the $\sin C + \sin D$ Sum-to-Product Formula: $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$.
$$\sin 6x + \sin 2x = 2 \sin \left(\frac{6x + 2x}{2}\right) \cos \left(\frac{6x – 2x}{2}\right)$$
$$= 2 \sin 4x \cos 2x$$
Substitute back into LHS:
$$\text{LHS} = 2 \sin 4x \cos 2x + 2 \sin 4x$$
Factor out $2 \sin 4x$:
$$\text{LHS} = 2 \sin 4x (\cos 2x + 1)$$
Use the Double Angle Identity $\cos 2x = 2 \cos^2 x – 1$, so $1 + \cos 2x = 2 \cos^2 x$:
$$\text{LHS} = 2 \sin 4x (2 \cos^2 x)$$
$$\text{LHS} = \mathbf{4 \cos^2 x \sin 4x} = \text{RHS}$$
15. Prove: $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x)$
LHS:
$$\text{LHS} = \cot 4x (\sin 5x + \sin 3x)$$
Use the $\sin C + \sin D$ Formula for the bracket:
$$\sin 5x + \sin 3x = 2 \sin \left(\frac{5x + 3x}{2}\right) \cos \left(\frac{5x – 3x}{2}\right) = 2 \sin 4x \cos x$$
$$\text{LHS} = \cot 4x (2 \sin 4x \cos x)$$
Substitute $\cot 4x = \frac{\cos 4x}{\sin 4x}$:
$$\text{LHS} = \frac{\cos 4x}{\sin 4x} (2 \sin 4x \cos x) = 2 \cos 4x \cos x$$
RHS:
$$\text{RHS} = \cot x (\sin 5x – \sin 3x)$$
Use the $\sin C – \sin D$ Sum-to-Product Formula: $\sin C – \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$.
$$\sin 5x – \sin 3x = 2 \cos \left(\frac{5x + 3x}{2}\right) \sin \left(\frac{5x – 3x}{2}\right) = 2 \cos 4x \sin x$$
$$\text{RHS} = \cot x (2 \cos 4x \sin x)$$
Substitute $\cot x = \frac{\cos x}{\sin x}$:
$$\text{RHS} = \frac{\cos x}{\sin x} (2 \cos 4x \sin x) = 2 \cos x \cos 4x$$
Since LHS = RHS ($2 \cos 4x \cos x$), the identity is proven.
16. Prove: $\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}$
We use the Sum-to-Product Formulas:
- Numerator ($\cos C – \cos D$): $\cos C – \cos D = -2 \sin \left(\frac{C + D}{2}\right) \sin \left(\frac{C – D}{2}\right)$
- Denominator ($\sin C – \sin D$): $\sin C – \sin D = 2 \cos \left(\frac{C + D}{2}\right) \sin \left(\frac{C – D}{2}\right)$
Numerator:
$$\cos 9x – \cos 5x = -2 \sin \left(\frac{9x + 5x}{2}\right) \sin \left(\frac{9x – 5x}{2}\right) = -2 \sin 7x \sin 2x$$
Denominator:
$$\sin 17x – \sin 3x = 2 \cos \left(\frac{17x + 3x}{2}\right) \sin \left(\frac{17x – 3x}{2}\right) = 2 \cos 10x \sin 7x$$
LHS:
$$\text{LHS} = \frac{-2 \sin 7x \sin 2x}{2 \cos 10x \sin 7x}$$
Cancel $2$ and $\sin 7x$ (assuming $\sin 7x \ne 0$):
$$\text{LHS} = \mathbf{-\frac{\sin 2x}{\cos 10x}} = \text{RHS}$$
17. Prove: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$
We use the Sum-to-Product Formulas:
- Numerator ($\sin C + \sin D$): $\sin C + \sin D = 2 \sin \left(\frac{C + D}{2}\right) \cos \left(\frac{C – D}{2}\right)$
- Denominator ($\cos C + \cos D$): $\cos C + \cos D = 2 \cos \left(\frac{C + D}{2}\right) \cos \left(\frac{C – D}{2}\right)$
Numerator:
$$\sin 5x + \sin 3x = 2 \sin \left(\frac{5x + 3x}{2}\right) \cos \left(\frac{5x – 3x}{2}\right) = 2 \sin 4x \cos x$$
Denominator:
$$\cos 5x + \cos 3x = 2 \cos \left(\frac{5x + 3x}{2}\right) \cos \left(\frac{5x – 3x}{2}\right) = 2 \cos 4x \cos x$$
LHS:
$$\text{LHS} = \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}$$
Cancel $2$ and $\cos x$ (assuming $\cos x \ne 0$):
$$\text{LHS} = \frac{\sin 4x}{\cos 4x} = \tan 4x = \text{RHS}$$
18. Prove: $\frac{\sin x – \sin y}{\cos x + \cos y} = \tan \left(\frac{x – y}{2}\right)$
We use the Sum-to-Product Formulas:
- Numerator ($\sin C – \sin D$): $\sin C – \sin D = 2 \cos \left(\frac{C + D}{2}\right) \sin \left(\frac{C – D}{2}\right)$
- Denominator ($\cos C + \cos D$): $\cos C + \cos D = 2 \cos \left(\frac{C + D}{2}\right) \cos \left(\frac{C – D}{2}\right)$
LHS:
$$\text{LHS} = \frac{2 \cos \left(\frac{x + y}{2}\right) \sin \left(\frac{x – y}{2}\right)}{2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x – y}{2}\right)}$$
Cancel $2$ and $\cos \left(\frac{x+y}{2}\right)$ (assuming denominator $\ne 0$):
$$\text{LHS} = \frac{\sin \left(\frac{x – y}{2}\right)}{\cos \left(\frac{x – y}{2}\right)} = \tan \left(\frac{x – y}{2}\right) = \text{RHS}$$
19. Prove: $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$
We use the Sum-to-Product Formulas (similar to Q.17):
- Numerator: $\sin x + \sin 3x = 2 \sin \left(\frac{x + 3x}{2}\right) \cos \left(\frac{x – 3x}{2}\right) = 2 \sin 2x \cos(-x) = 2 \sin 2x \cos x$
- Denominator: $\cos x + \cos 3x = 2 \cos \left(\frac{x + 3x}{2}\right) \cos \left(\frac{x – 3x}{2}\right) = 2 \cos 2x \cos(-x) = 2 \cos 2x \cos x$
LHS:
$$\text{LHS} = \frac{2 \sin 2x \cos x}{2 \cos 2x \cos x}$$
Cancel $2$ and $\cos x$:
$$\text{LHS} = \frac{\sin 2x}{\cos 2x} = \tan 2x = \text{RHS}$$
20. Prove: $\frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x} = 2 \sin x$
We use the Sum-to-Product Formula for the numerator and the Double Angle Identity for the denominator.
Numerator ($\sin C – \sin D$):
$$\sin x – \sin 3x = 2 \cos \left(\frac{x + 3x}{2}\right) \sin \left(\frac{x – 3x}{2}\right)$$
$$= 2 \cos 2x \sin (-x) = -2 \cos 2x \sin x$$
Denominator:
$$\sin^2 x – \cos^2 x = -(\cos^2 x – \sin^2 x)$$
Using $\cos 2x = \cos^2 x – \sin^2 x$:
$$\sin^2 x – \cos^2 x = -\cos 2x$$
LHS:
$$\text{LHS} = \frac{-2 \cos 2x \sin x}{-\cos 2x}$$
Cancel $-\cos 2x$ (assuming denominator $\ne 0$):
$$\text{LHS} = \mathbf{2 \sin x} = \text{RHS}$$
21. Prove: $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$
We group the terms in the numerator and denominator to use the sum-to-product formulas.
Numerator:
$$\text{Num} = (\cos 4x + \cos 2x) + \cos 3x$$
Use $\cos C + \cos D = 2 \cos \left(\frac{C + D}{2}\right) \cos \left(\frac{C – D}{2}\right)$:
$$\cos 4x + \cos 2x = 2 \cos 3x \cos x$$
$$\text{Num} = 2 \cos 3x \cos x + \cos 3x$$
Factor out $\cos 3x$:
$$\text{Num} = \cos 3x (2 \cos x + 1)$$
Denominator:
$$\text{Den} = (\sin 4x + \sin 2x) + \sin 3x$$
Use $\sin C + \sin D = 2 \sin \left(\frac{C + D}{2}\right) \cos \left(\frac{C – D}{2}\right)$:
$$\sin 4x + \sin 2x = 2 \sin 3x \cos x$$
$$\text{Den} = 2 \sin 3x \cos x + \sin 3x$$
Factor out $\sin 3x$:
$$\text{Den} = \sin 3x (2 \cos x + 1)$$
LHS:
$$\text{LHS} = \frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)}$$
Cancel $(2 \cos x + 1)$:
$$\text{LHS} = \frac{\cos 3x}{\sin 3x} = \cot 3x = \text{RHS}$$
22. Prove: $\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$
The angles are $x, 2x, 3x$. Notice that $3x = 2x + x$.
Start with the tangent of the sum formula for $3x$:
$$\tan 3x = \tan(2x + x) = \frac{\tan 2x + \tan x}{1 – \tan 2x \tan x}$$
Cross-multiply:
$$\tan 3x (1 – \tan 2x \tan x) = \tan 2x + \tan x$$
$$\tan 3x – \tan 3x \tan 2x \tan x = \tan 2x + \tan x$$
Rearrange the terms:
$$\tan 3x – \tan 2x – \tan x = \tan 3x \tan 2x \tan x$$
Divide the entire equation by $\tan x \tan 2x \tan 3x$:
$$\frac{\tan 3x}{\tan x \tan 2x \tan 3x} – \frac{\tan 2x}{\tan x \tan 2x \tan 3x} – \frac{\tan x}{\tan x \tan 2x \tan 3x} = 1$$
Simplify (using $\frac{1}{\tan \theta} = \cot \theta$):
$$\frac{1}{\tan x \tan 2x} – \frac{1}{\tan x \tan 3x} – \frac{1}{\tan 2x \tan 3x} = 1$$
$$\cot x \cot 2x – \cot x \cot 3x – \cot 2x \cot 3x = 1$$
This is equivalent to the required identity:
$$\mathbf{\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1}$$
23. Prove: $\frac{\sin 4x}{\sin x} = 4 \cos x (1 – 2 \sin^2 x)$
(Note: The question provided $\frac{\sin 4x}{\sin x} = \frac{4\tan x (1-\tan^2 x)}{1-6\tan^2 x+\tan^4 x}$ which is the formula for $\tan 4x$. Assuming the question intended to ask for $\tan 4x$ or $\cos 4x$ or had a typo. Let’s solve $\tan 4x$ using the given RHS in the exercise text for completeness and show it simplifies to $\frac{\sin 4x}{\cos 4x}$).
Assuming the question is to prove the correct expansion of $\tan 4x$:
$$\frac{4\tan x (1-\tan^2 x)}{1-6\tan^2 x+\tan^4 x} = \tan 4x$$
We use the Double Angle Formula for $\tan 2x$: $\tan 2x = \frac{2 \tan x}{1 – \tan^2 x}$.
$$\tan 4x = \tan (2 \cdot 2x) = \frac{2 \tan 2x}{1 – \tan^2 2x}$$
Substitute the expression for $\tan 2x$:
$$\tan 4x = \frac{2 \left(\frac{2 \tan x}{1 – \tan^2 x}\right)}{1 – \left(\frac{2 \tan x}{1 – \tan^2 x}\right)^2}$$
$$\tan 4x = \frac{\frac{4 \tan x}{1 – \tan^2 x}}{1 – \frac{4 \tan^2 x}{(1 – \tan^2 x)^2}}$$
Find a common denominator in the denominator:
$$\tan 4x = \frac{\frac{4 \tan x}{1 – \tan^2 x}}{\frac{(1 – \tan^2 x)^2 – 4 \tan^2 x}{(1 – \tan^2 x)^2}}$$
Invert and multiply:
$$\tan 4x = \frac{4 \tan x}{1 – \tan^2 x} \times \frac{(1 – \tan^2 x)^2}{1 – 2 \tan^2 x + \tan^4 x – 4 \tan^2 x}$$
Cancel one factor of $(1 – \tan^2 x)$:
$$\tan 4x = \frac{4 \tan x (1 – \tan^2 x)}{1 – 6 \tan^2 x + \tan^4 x}$$
$$\text{LHS} = \text{RHS}$$
24. Prove: $\cos 4x = 1 – 8 \sin^2 x \cos^2 x$
We use the Double Angle Identities.
Start with $\cos 4x = \cos (2 \cdot 2x)$.
Use the form $\cos 2\theta = 1 – 2 \sin^2 \theta$, where $\theta = 2x$:
$$\cos 4x = 1 – 2 \sin^2 (2x)$$
Now use the identity $\sin 2x = 2 \sin x \cos x$:
$$\cos 4x = 1 – 2 (\sin 2x)^2$$
$$\cos 4x = 1 – 2 (2 \sin x \cos x)^2$$
$$\cos 4x = 1 – 2 (4 \sin^2 x \cos^2 x)$$
$$\cos 4x = \mathbf{1 – 8 \sin^2 x \cos^2 x} = \text{RHS}$$
25. Prove: $\cos 6x = 32 \cos^6 x – 48 \cos^4 x + 18 \cos^2 x – 1$
We use the Triple Angle and Double Angle Identities.
Start with $\cos 6x = \cos(3 \cdot 2x)$.
Use the identity $\cos 3\theta = 4 \cos^3 \theta – 3 \cos \theta$, where $\theta = 2x$:
$$\cos 6x = 4 \cos^3 (2x) – 3 \cos (2x)$$
Now substitute the identity $\cos 2x = 2 \cos^2 x – 1$:
$$\cos 6x = 4 (2 \cos^2 x – 1)^3 – 3 (2 \cos^2 x – 1)$$
Expand the cubic term using $(A-B)^3 = A^3 – 3A^2 B + 3AB^2 – B^3$:
$$\cos 6x = 4 [ (2 \cos^2 x)^3 – 3(2 \cos^2 x)^2(1) + 3(2 \cos^2 x)(1)^2 – 1^3 ] – 6 \cos^2 x + 3$$
$$\cos 6x = 4 [ 8 \cos^6 x – 12 \cos^4 x + 6 \cos^2 x – 1 ] – 6 \cos^2 x + 3$$
Distribute the 4:
$$\cos 6x = 32 \cos^6 x – 48 \cos^4 x + 24 \cos^2 x – 4 – 6 \cos^2 x + 3$$
Combine like terms:
$$\cos 6x = \mathbf{32 \cos^6 x – 48 \cos^4 x + 18 \cos^2 x – 1} = \text{RHS}$$
