RBSE Solutions Class 11 Maths Chapter 3 त्रिकोणमितीय फलन Exercise 3.3

सिद्ध कीजिए

Question 1. \( \sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2} \)
Answer:
प्रश्नानुसार L.H.S.
\( = \sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} \)
\( = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 - (1)^2 \)
\( = \frac{1}{4} + \frac{1}{4} - 1 \)
\( = \frac{1+1-4}{4} \)
\( = \frac{-2}{4} \)
\( = -\frac{1}{2} = \text{R.H.S.} \)
In simple words: We replace the trigonometric functions with their known values for the given angles, square them, and then add and subtract to show that the left side equals the right side, which is \( -\frac{1}{2} \).

🎯 Exam Tip: Remember the exact values of trigonometric functions for standard angles like \( \frac{\pi}{6} \) (30°), \( \frac{\pi}{3} \) (60°), and \( \frac{\pi}{4} \) (45°). A common mistake is using incorrect signs or values.

Question 2. \( 2 \sin^2 \frac{\pi}{6} + \text{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2} \)
Answer:
L.H.S. \( = 2\sin^2 \frac{\pi}{6} + \text{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} \)
We know that \( \text{cosec} \frac{7\pi}{6} = \text{cosec} \left( \pi + \frac{\pi}{6} \right) \)
\( = -\text{cosec} \frac{\pi}{6} \)
\( = -2 \)
So, L.H.S. \( = 2\sin^2 \frac{\pi}{6} + \left( -\text{cosec} \frac{\pi}{6} \right)^2 \cos^2 \frac{\pi}{3} \)
\( = 2 \left( \frac{1}{2} \right)^2 + (-2)^2 \left( \frac{1}{2} \right)^2 \)
\( = 2 \times \frac{1}{4} + 4 \times \frac{1}{4} \)
\( = \frac{1}{2} + 1 \)
\( = \frac{3}{2} = \text{R.H.S.} \)
In simple words: First, we simplify the cosecant term by using its periodicity. Then, we substitute the known values for sine, cosecant, and cosine at the given angles and calculate the expression to show it equals \( \frac{3}{2} \).

🎯 Exam Tip: When dealing with angles greater than \( \frac{\pi}{2} \), always reduce them to standard angles using quadrant rules (e.g., \( \sin(\pi + \theta) = -\sin \theta \)) to find their values.

Question 3. \( \cot^2 \frac{\pi}{6} + \text{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6 \)
Answer:
L.H.S. \( = \cot^2 \frac{\pi}{6} + \text{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} \)
We know that \( \text{cosec} \frac{5\pi}{6} = \text{cosec} \left( \pi - \frac{\pi}{6} \right) \)
\( = \text{cosec} \frac{\pi}{6} \)
\( = 2 \)
So, L.H.S. \( = (\sqrt{3})^2 + 2 + 3 \left( \frac{1}{\sqrt{3}} \right)^2 \)
\( = 3 + 2 + 3 \left( \frac{1}{3} \right) \)
\( = 3 + 2 + 1 \)
\( = 6 = \text{R.H.S.} \)
In simple words: We first simplify the cosecant term using angle properties. Then, we plug in the exact values of cotangent, cosecant, and tangent for the given angles, perform the calculations, and verify that the result is 6.

🎯 Exam Tip: Pay close attention to the signs of trigonometric functions in different quadrants. For \( \text{cosec} \frac{5\pi}{6} \), it's in the second quadrant where cosecant is positive.

Question 4. \( 2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10 \)
Answer:
L.H.S. \( = 2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} \)
We know that \( \sin \frac{3\pi}{4} = \sin \left( \pi - \frac{\pi}{4} \right) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
So, L.H.S. \( = 2 \left( \frac{1}{\sqrt{2}} \right)^2 + 2 \left( \frac{1}{\sqrt{2}} \right)^2 + 2 (2)^2 \)
\( = 2 \left( \frac{1}{2} \right) + 2 \left( \frac{1}{2} \right) + 2(4) \)
\( = 1 + 1 + 8 \)
\( = 10 = \text{R.H.S.} \)
In simple words: We find the values of sine, cosine, and secant for the given angles. Then, we substitute these values into the expression, square them, and add them up to prove that the total equals 10.

🎯 Exam Tip: Remember that \( \sec \theta = \frac{1}{\cos \theta} \). If you know \( \cos \frac{\pi}{3} = \frac{1}{2} \), then \( \sec \frac{\pi}{3} \) is its reciprocal, which is 2.

Question 5. मान ज्ञात कीजिया
(i) sin 75°
(ii) tan 15°
Answer:
(i) We need to find the value of \( \sin 75^\circ \).
\( \sin 75^\circ = \sin (45^\circ + 30^\circ) \)
Using the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \):
\( = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) \)
\( = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
(ii) We need to find the value of \( \tan 15^\circ \).
\( \tan 15^\circ = \tan (45^\circ - 30^\circ) \)
Using the identity \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):
\( = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \)
\( = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1) \frac{1}{\sqrt{3}}} \)
\( = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \)
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)
To rationalize the denominator, multiply the numerator and denominator by \( (\sqrt{3}-1) \):
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - (1)^2} \)
\( = \frac{(\sqrt{3})^2 - 2\sqrt{3} + (1)^2}{3 - 1} \)
\( = \frac{3 - 2\sqrt{3} + 1}{2} \)
\( = \frac{4 - 2\sqrt{3}}{2} \)
\( = 2 - \sqrt{3} \)
In simple words: For sin 75°, we use the sum formula for sine with 45° and 30°. For tan 15°, we use the difference formula for tangent with 45° and 30°, then simplify the result by rationalizing the denominator.

🎯 Exam Tip: Memorize the angle sum and difference formulas for sine, cosine, and tangent, as they are essential for solving such problems. Rationalize denominators for cleaner final answers.

Question 6. सिद्ध कीजिए
\( \tan 225^\circ \cot 405^\circ + \tan 765^\circ \cot 675^\circ = 0 \)
Answer:
L.H.S. \( = \tan 225^\circ \cot 405^\circ + \tan 765^\circ \cot 675^\circ \)
Let's simplify each term:
\( \tan 225^\circ = \tan(180^\circ + 45^\circ) = \tan 45^\circ = 1 \)
\( \cot 405^\circ = \cot(360^\circ + 45^\circ) = \cot 45^\circ = 1 \)
\( \tan 765^\circ = \tan(2 \times 360^\circ + 45^\circ) = \tan 45^\circ = 1 \)
\( \cot 675^\circ = \cot(2 \times 360^\circ - 45^\circ) = \cot(-45^\circ) = -\cot 45^\circ = -1 \)
Now substitute these values into the L.H.S.:
L.H.S. \( = (1)(1) + (1)(-1) \)
\( = 1 - 1 \)
\( = 0 = \text{R.H.S.} \)
In simple words: We simplify each trigonometric term by finding its equivalent value for angles within 0° to 360°. After simplification, we substitute these values back into the expression and find that the sum equals zero.

🎯 Exam Tip: Use the periodicity of trigonometric functions (e.g., \( \tan(n \times 180^\circ + \theta) = \tan \theta \), \( \cot(n \times 180^\circ + \theta) = \cot \theta \)) to reduce large angles. Also, remember \( \cot(-\theta) = -\cot \theta \).

Question 8. निम्नलिखित को सिद्ध कीजिए
\( \cos \left( \frac{\pi}{4} - x \right) \cos \left( \frac{\pi}{4} - y \right) - \sin \left( \frac{\pi}{4} - x \right) \sin \left( \frac{\pi}{4} - y \right) = \sin (x + y) \)
Answer:
L.H.S. \( = \cos \left( \frac{\pi}{4} - x \right) \cos \left( \frac{\pi}{4} - y \right) - \sin \left( \frac{\pi}{4} - x \right) \sin \left( \frac{\pi}{4} - y \right) \)
Let \( A = \frac{\pi}{4} - x \) and \( B = \frac{\pi}{4} - y \).
The expression becomes \( \cos A \cos B - \sin A \sin B \).
This is the formula for \( \cos(A+B) \).
So, L.H.S. \( = \cos \left( \left( \frac{\pi}{4} - x \right) + \left( \frac{\pi}{4} - y \right) \right) \)
\( = \cos \left( \frac{\pi}{4} - x + \frac{\pi}{4} - y \right) \)
\( = \cos \left( \frac{2\pi}{4} - (x+y) \right) \)
\( = \cos \left( \frac{\pi}{2} - (x+y) \right) \)
Using the identity \( \cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta \):
\( = \sin (x+y) = \text{R.H.S.} \)
In simple words: We use the cosine addition formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). By setting \( A = \frac{\pi}{4} - x \) and \( B = \frac{\pi}{4} - y \), the expression simplifies to \( \cos(\frac{\pi}{2} - (x+y)) \), which is equal to \( \sin(x+y) \).

🎯 Exam Tip: Recognize standard trigonometric formulas quickly. The structure of the expression directly points to the cosine addition formula. Then, apply co-function identities like \( \cos(\frac{\pi}{2} - \theta) = \sin \theta \).

Question 10. \( \frac{\cos(\pi+x) \cos(-x)}{\sin(\pi-x) \cos(\frac{\pi}{2}+x)} = \cot^2x \)
Answer:
L.H.S. \( = \frac{\cos(\pi+x) \cos(-x)}{\sin(\pi-x) \cos(\frac{\pi}{2}+x)} \)
We use the following trigonometric identities:
\( \cos(\pi+x) = -\cos x \)
\( \cos(-x) = \cos x \)
\( \sin(\pi-x) = \sin x \)
\( \cos\left(\frac{\pi}{2}+x\right) = -\sin x \)
Substitute these into the L.H.S.:
\( = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \)
\( = \frac{-\cos^2 x}{-\sin^2 x} \)
\( = \frac{\cos^2 x}{\sin^2 x} \)
\( = \cot^2 x = \text{R.H.S.} \)
In simple words: We replace each trigonometric term with its equivalent simplified form using quadrant rules. Then, we multiply and divide the terms to simplify the expression, showing that it equals \( \cot^2x \).

🎯 Exam Tip: Clearly remember the sign conventions and transformations for trigonometric functions in different quadrants and for negative angles (e.g., \( \cos(-x) = \cos x \), \( \cos(\pi+x) = -\cos x \)).

Question 11. \( \sin (n + 1) x \sin (n + 2) x + \cos (n + 1) x \cos (n + 2) x = \cos x \)
Answer:
L.H.S. \( = \sin (n + 1) x \sin (n + 2) x + \cos (n + 1) x \cos (n + 2) x \)
Let \( A = (n + 1) x \) and \( B = (n + 2) x \).
The expression matches the formula for \( \cos(A-B) \), which is \( \cos A \cos B + \sin A \sin B \).
So, L.H.S. \( = \cos[(n + 1) x - (n + 2) x] \)
\( = \cos[nx + x - nx - 2x] \)
\( = \cos[x - 2x] \)
\( = \cos(-x) \)
We know that \( \cos(-\theta) = \cos \theta \) for all \( \theta \in \mathbb{R} \).
\( = \cos x = \text{R.H.S.} \)
In simple words: We recognize that the given expression is in the form of the cosine difference formula. By substituting the terms, the expression simplifies to \( \cos(-x) \), which is equal to \( \cos x \).

🎯 Exam Tip: Identify the correct trigonometric identity to use. In this case, the `cos A cos B + sin A sin B` pattern is a clear indicator of \( \cos(A-B) \).

Question 12. \( \sin^2 6x - \sin^2 4x = \sin 2x \sin 10x \)
Answer:
L.H.S. \( = \sin^2 6x - \sin^2 4x \)
Using the identity \( a^2 - b^2 = (a+b)(a-b) \):
\( = (\sin 6x + \sin 4x)(\sin 6x - \sin 4x) \)
Now apply the sum and difference formulas for sine:
\( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \)
\( \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \)
So, \( (\sin 6x + \sin 4x) = 2 \sin \frac{6x+4x}{2} \cos \frac{6x-4x}{2} = 2 \sin 5x \cos x \)
And \( (\sin 6x - \sin 4x) = 2 \cos \frac{6x+4x}{2} \sin \frac{6x-4x}{2} = 2 \cos 5x \sin x \)
Substitute these back into the L.H.S.:
\( = (2 \sin 5x \cos x)(2 \cos 5x \sin x) \)
\( = (2 \sin 5x \cos 5x)(2 \sin x \cos x) \)
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\( = (\sin (2 \times 5x))(\sin (2 \times x)) \)
\( = \sin 10x \sin 2x \)
\( = \sin 2x \sin 10x = \text{R.H.S.} \)
In simple words: We use the difference of squares formula and then apply the sum-to-product and product-to-sum trigonometric identities. This allows us to simplify the expression into a product of sines, matching the right side of the equation.

🎯 Exam Tip: The identity \( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \) can directly simplify this. Alternatively, applying sum-to-product formulas after difference of squares works as shown.

Question 13. \( \sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x \)
Answer:
L.H.S. \( = \sin 2x + 2 \sin 4x + \sin 6x \)
Rearrange the terms:
\( = (\sin 6x + \sin 2x) + 2 \sin 4x \)
Apply the sum formula for sine, \( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \):
\( = 2 \sin \frac{6x+2x}{2} \cos \frac{6x-2x}{2} + 2 \sin 4x \)
\( = 2 \sin 4x \cos 2x + 2 \sin 4x \)
Factor out \( 2 \sin 4x \):
\( = 2 \sin 4x (\cos 2x + 1) \)
We know the double angle identity \( \cos 2x = 2 \cos^2 x - 1 \), so \( \cos 2x + 1 = 2 \cos^2 x \).
\( = 2 \sin 4x (2 \cos^2 x) \)
\( = 4 \cos^2 x \sin 4x = \text{R.H.S.} \)
In simple words: We group the sine terms, apply the sum-to-product formula, and then factor out the common term. Finally, we use a double angle identity for cosine to simplify the expression and match the right side.

🎯 Exam Tip: Look for opportunities to group terms and apply sum-to-product formulas. Also, be ready to use variations of double angle identities (like \( \cos 2x + 1 = 2\cos^2 x \)) to simplify expressions.

Question 14. \( \cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x) \)
Answer:
Let's simplify the L.H.S. and R.H.S. separately.
L.H.S. \( = \cot 4x (\sin 5x + \sin 3x) \)
Apply \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( = \frac{\cos 4x}{\sin 4x} \left( 2 \sin \frac{5x+3x}{2} \cos \frac{5x-3x}{2} \right) \)
\( = \frac{\cos 4x}{\sin 4x} (2 \sin 4x \cos x) \)
\( = 2 \cos 4x \cos x \) (i)
Now, R.H.S. \( = \cot x (\sin 5x - \sin 3x) \)
Apply \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \):
\( = \frac{\cos x}{\sin x} \left( 2 \cos \frac{5x+3x}{2} \sin \frac{5x-3x}{2} \right) \)
\( = \frac{\cos x}{\sin x} (2 \cos 4x \sin x) \)
\( = 2 \cos x \cos 4x \) (ii)
From (i) and (ii), L.H.S. \( = \) R.H.S.
In simple words: We simplify both sides of the equation separately. For the left side, we use the sum-to-product formula for sine. For the right side, we use the difference-to-product formula for sine. Both sides simplify to the same expression, proving the identity.

🎯 Exam Tip: When proving identities involving products and sums, it's often effective to simplify both sides independently until they match. Remember to convert cotangent to \( \frac{\cos}{\sin} \).

Question 15. सिद्ध कीजिए
\( \frac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} = \tan x \)
Answer:
L.H.S. \( = \frac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} \)
Rearrange the numerator:
\( = \frac{(\sin 5x + \sin x) - 2 \sin 3x}{\cos 5x - \cos x} \)
Apply sum-to-product formula for numerator \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( (\sin 5x + \sin x) = 2 \sin \frac{5x+x}{2} \cos \frac{5x-x}{2} = 2 \sin 3x \cos 2x \)
Apply difference-to-product formula for denominator \( \cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2} \):
\( (\cos 5x - \cos x) = -2 \sin \frac{5x+x}{2} \sin \frac{5x-x}{2} = -2 \sin 3x \sin 2x \)
Substitute these back into the L.H.S.:
\( = \frac{2 \sin 3x \cos 2x - 2 \sin 3x}{-2 \sin 3x \sin 2x} \)
Factor out \( 2 \sin 3x \) from the numerator:
\( = \frac{2 \sin 3x (\cos 2x - 1)}{-2 \sin 3x \sin 2x} \)
Cancel \( 2 \sin 3x \) (assuming \( \sin 3x \neq 0 \)):
\( = \frac{\cos 2x - 1}{-\sin 2x} \)
We know that \( \cos 2x - 1 = -2 \sin^2 x \), and \( \sin 2x = 2 \sin x \cos x \).
So, \( = \frac{-2 \sin^2 x}{-(2 \sin x \cos x)} \)
\( = \frac{2 \sin^2 x}{2 \sin x \cos x} \)
\( = \frac{\sin x}{\cos x} \)
\( = \tan x = \text{R.H.S.} \)
In simple words: We first rearrange the numerator, then use sum-to-product and difference-to-product formulas for the numerator and denominator, respectively. After simplifying by canceling common terms, we apply double angle identities for cosine and sine to reach \( \tan x \).

🎯 Exam Tip: Look for opportunities to group terms like \( \sin 5x + \sin x \) to apply sum-to-product formulas. Remember that \( \cos 2x - 1 = -2 \sin^2 x \) is a useful variant of the double angle formula.

Question 16. \( \frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x-y}{2} \)
Answer:
L.H.S. \( = \frac{\sin x - \sin y}{\cos x + \cos y} \)
Apply difference-to-product formula for numerator \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \):
\( \sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \)
Apply sum-to-product formula for denominator \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( \cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} \)
Substitute these back into the L.H.S.:
\( = \frac{2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}} \)
Cancel the common term \( 2 \cos \frac{x+y}{2} \) (assuming it's not zero):
\( = \frac{\sin \frac{x-y}{2}}{\cos \frac{x-y}{2}} \)
\( = \tan \frac{x-y}{2} = \text{R.H.S.} \)
In simple words: We use the difference-to-product formula for the sine terms in the numerator and the sum-to-product formula for the cosine terms in the denominator. After canceling the common factor, the expression simplifies to \( \tan \frac{x-y}{2} \).

🎯 Exam Tip: Directly apply the sum-to-product and difference-to-product formulas. Make sure to remember the correct formula for each case, especially the sign for \( \cos C - \cos D \).

Question 17. \( \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x \)
Answer:
L.H.S. \( = \frac{\sin x + \sin 3x}{\cos x + \cos 3x} \)
Apply sum-to-product formula for numerator \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( \sin x + \sin 3x = 2 \sin \frac{x+3x}{2} \cos \frac{x-3x}{2} \)
\( = 2 \sin 2x \cos (-x) \)
Apply sum-to-product formula for denominator \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( \cos x + \cos 3x = 2 \cos \frac{x+3x}{2} \cos \frac{x-3x}{2} \)
\( = 2 \cos 2x \cos (-x) \)
Substitute these back into the L.H.S.:
\( = \frac{2 \sin 2x \cos (-x)}{2 \cos 2x \cos (-x)} \)
Cancel the common term \( 2 \cos (-x) \) (assuming it's not zero):
\( = \frac{\sin 2x}{\cos 2x} \)
\( = \tan 2x = \text{R.H.S.} \)
In simple words: We apply the sum-to-product formulas for both the numerator and the denominator. After simplifying by canceling the common terms and knowing that \( \cos(-x) = \cos x \), the expression reduces to \( \tan 2x \).

🎯 Exam Tip: Remember that \( \cos(-\theta) = \cos \theta \). This is crucial for correctly simplifying terms like \( \cos(-x) \) to \( \cos x \). Always look for common factors to cancel after applying identities.

Question 19. \( \tan 4x = \frac{4 \tan x (1-\tan^2x)}{1-6\tan^2x+\tan^4x} \)
Answer:
L.H.S. \( = \tan 4x \)
We can write \( \tan 4x \) as \( \tan (2 \times 2x) \).
Using the double angle formula \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \):
\( = \frac{2 \tan 2x}{1 - \tan^2 2x} \)
Now substitute the formula for \( \tan 2x \) again into this expression:
\( = \frac{2 \left( \frac{2 \tan x}{1 - \tan^2 x} \right)}{1 - \left( \frac{2 \tan x}{1 - \tan^2 x} \right)^2} \)
\( = \frac{\frac{4 \tan x}{1 - \tan^2 x}}{1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}} \)
Find a common denominator for the denominator's terms:
\( = \frac{\frac{4 \tan x}{1 - \tan^2 x}}{\frac{(1 - \tan^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \frac{4 \tan x}{1 - \tan^2 x} \times \frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2 - 4 \tan^2 x} \)
Cancel one \( (1 - \tan^2 x) \) term:
\( = \frac{4 \tan x (1 - \tan^2 x)}{(1 - 2\tan^2 x + \tan^4 x) - 4 \tan^2 x} \)
\( = \frac{4 \tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x} = \text{R.H.S.} \)
In simple words: We express \( \tan 4x \) as \( \tan(2 \times 2x) \) and apply the double angle formula for tangent twice. After substituting the formula, we simplify the complex fraction by finding a common denominator and expanding the terms to match the right side of the equation.

🎯 Exam Tip: When dealing with \( \tan 4x \) or similar multiple angles, break them down into double angles (e.g., \( \tan(2 \times 2x) \)). Be careful with algebraic simplification of complex fractions and expanding squared terms.

Question 20. \( \cos 4x = 1 - 8\sin^2 x \cos^2 x \)
Answer:
L.H.S. \( = \cos 4x \)
We can write \( \cos 4x \) as \( \cos(2 \times 2x) \).
Using the double angle formula \( \cos 2\theta = 1 - 2\sin^2 \theta \):
\( = 1 - 2\sin^2 (2x) \)
We know that \( \sin 2x = 2 \sin x \cos x \). Substitute this:
\( = 1 - 2(2 \sin x \cos x)^2 \)
\( = 1 - 2(4 \sin^2 x \cos^2 x) \)
\( = 1 - 8\sin^2 x \cos^2 x = \text{R.H.S.} \)
In simple words: We rewrite \( \cos 4x \) as \( \cos(2 \times 2x) \) and use the double angle formula \( \cos 2\theta = 1 - 2\sin^2 \theta \). Then, we substitute the formula for \( \sin 2x \) and simplify the expression to get \( 1 - 8\sin^2 x \cos^2 x \).

🎯 Exam Tip: When an identity involves \( \cos 4x \), usually the first step is to apply a double angle formula twice. Choose the appropriate version of the \( \cos 2\theta \) formula that leads towards the target expression, in this case, the one involving \( \sin^2 \theta \).

Question 21. \( \cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \)
Answer:
L.H.S. \( = \cos 6x \)
We can write \( \cos 6x \) as \( \cos(2 \times 3x) \).
Using the double angle formula \( \cos 2\theta = 2\cos^2 \theta - 1 \):
\( = 2\cos^2 (3x) - 1 \)
Now, use the triple angle formula \( \cos 3x = 4\cos^3 x - 3\cos x \). Substitute this into the expression:
\( = 2(4\cos^3 x - 3\cos x)^2 - 1 \)
Expand the squared term \( (A-B)^2 = A^2 - 2AB + B^2 \):
\( = 2[(4\cos^3 x)^2 - 2(4\cos^3 x)(3\cos x) + (3\cos x)^2] - 1 \)
\( = 2[16\cos^6 x - 24\cos^4 x + 9\cos^2 x] - 1 \)
Distribute the 2:
\( = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1 = \text{R.H.S.} \)
In simple words: We rewrite \( \cos 6x \) as \( \cos(2 \times 3x) \) and apply the double angle formula for cosine. Then, we substitute the triple angle formula for \( \cos 3x \) and expand the squared expression carefully. Finally, we simplify the terms to match the complex polynomial expression on the right side.

🎯 Exam Tip: This type of question requires memorizing and correctly applying both double and triple angle formulas. Be very careful with the algebraic expansion of squared terms, especially with higher powers of cosine.