CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 06

Question. Express \(\sin 67^\circ + \cos 75^\circ\) in terms of trigonometric ratios of angles between \(0^\circ\) and \(45^\circ\).
Answer: Solution. \(\sin 67^\circ + \cos 75^\circ = \sin(90^\circ - 23^\circ) + \cos(90^\circ - 15^\circ) = \cos 23^\circ + \sin 15^\circ \)

Question. If \(\sin 3\theta = \cos(\theta - 6^\circ)\), where \(3\theta\) and \((\theta - 6^\circ)\) are acute angles, find the value of \(\theta\).
Answer: Solution. \(\sin 3\theta = \cos (\theta - 6^\circ) \)
\( \implies \) \( \cos (90^\circ - 3\theta) = \cos (\theta - 6^\circ) \)
\( \implies \) \( 90^\circ - 3\theta = \theta - 6^\circ \)
\( \implies \) \( -3\theta - \theta = -6^\circ - 90^\circ \)
\( \implies \) \( 4\theta = 96^\circ \)
\( \implies \) \( \theta = 24^\circ \).

Question. If \(\tan 2\theta = \cot (\theta + 18^\circ)\), where \(2\theta\) and \((\theta + 18^\circ)\) are acute angles, find the value of \(\theta\).
Answer: Solution. \(\tan 2\theta = \cot (\theta + 18^\circ) \)
\( \implies \) \( \cot (90^\circ - 2\theta) = \cot (\theta + 18^\circ) \)
\( \implies \) \( 90^\circ - 2\theta = \theta + 18^\circ \)
\( \implies \) \( 2\theta + \theta = 90^\circ - 18^\circ \)
\( \implies \) \( 3\theta = 72^\circ \)
\( \implies \) \( \theta = \frac{72^\circ}{3} = 24^\circ \)

Question. If \(\sin (\theta + 24^\circ) = \cos \theta\), where \((\theta + 24^\circ)\) is an acute angle. Then find the value of \(\theta\).
Answer: Solution. \(\sin (\theta + 24^\circ) = \cos \theta \)
\( \implies \) \( \sin (\theta + 24^\circ) = \sin (90^\circ - \theta) \)
\( \implies \) \( \theta + \theta = 90^\circ - 24^\circ \)
\( \implies \) \( 2\theta = 66^\circ \)
\( \implies \) \( \theta + 24^\circ = 90^\circ - \theta \)
\( \implies \) \( \theta = \frac{66^\circ}{2} = 33^\circ \)

Verify that:

Question. \(\cos 60^\circ = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1}{2}\)
Answer: Solution. L.H.S. = \(\cos 60^\circ = \frac{1}{2}\)
R.H.S. = \(\frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \left(\frac{1}{\sqrt{3}}\right)^2}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{4} = \frac{1}{2} = \text{L.H.S.}\)
\( \implies \) \(\cos 60^\circ = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1}{2}\). Verified.

Question. \(\cos 60^\circ = \cos^2 30^\circ - \sin^2 30^\circ = \frac{1}{2}\)
Answer: Solution. L.H.S. = \(\cos 60^\circ = \frac{1}{2}\)
R.H.S. = \(\cos^2 30^\circ - \sin^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} = \text{L.H.S.}\)
\( \implies \) \(\cos 60^\circ = \cos^2 30^\circ - \sin^2 30^\circ = \frac{1}{2}\). Verified.

Question. For \(A = 30^\circ\), verify that: \(\cos 3A = 4\cos^3 A - 3\cos A\)
Answer: Solution. L.H.S. = \(\cos 3A = \cos 3 (30^\circ) = \cos 90^\circ = 0\)
R.H.S. = \(4 \cos^3 A - 3 \cos A\)
\( = 4 \cos^3 30^\circ - 3 \cos 30^\circ = 4 \left( \frac{\sqrt{3}}{2} \right)^3 - 3 \left( \frac{\sqrt{3}}{2} \right) \)
\( = \frac{4 \times 3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 = \text{L.H.S.}\). Verified.

Question. Verify that \(\frac{\cot 30^\circ \cot 60^\circ - 1}{\cot 30^\circ + \cot 60^\circ} = \cot 90^\circ\)
Answer: Solution. L.H.S. = \(\frac{\cot 30^\circ \cot 60^\circ - 1}{\cot 30^\circ + \cot 60^\circ} = \frac{\sqrt{3} \times \frac{1}{\sqrt{3}} - 1}{\sqrt{3} + \frac{1}{\sqrt{3}}} = \frac{1 - 1}{\frac{3 + 1}{\sqrt{3}}} = 0 \times \frac{\sqrt{3}}{4} = 0 \)
R.H.S. = \(\cot 90^\circ = 0 = \text{L.H.S.}\)
\( \implies \) \(\frac{\cot 30^\circ \cot 60^\circ - 1}{\cot 30^\circ + \cot 60^\circ} = \cot 90^\circ\). Verified.

Question. If \(\angle A = \angle B = 45^\circ\), verify that: \(\sin (A + B) = \sin A \cos B + \cos A \sin B\).
Answer: Solution. \(\angle A = \angle B = 45^\circ\)
L.H.S. = \(\sin (A + B) = \sin (45^\circ + 45^\circ) = \sin 90^\circ = 1\)
R.H.S. = \(\sin A \cos B + \cos A \sin B\)
\( = \sin 45^\circ \cos 45^\circ + \cos 45^\circ \sin 45^\circ \)
\( = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1 = \text{L.H.S.}\). Verified.

Question. Using the formula \(\sin \theta = \sqrt{\frac{1 - \cos 2\theta}{2}}\), find the value of \(\sin 30^\circ\), given that \(\cos 60^\circ = \frac{1}{2}\).
Answer: Solution. \(\sin \theta = \sqrt{\frac{1 - \cos 2\theta}{2}}\); Given \(\cos 60^\circ = \frac{1}{2}\)
Put \(\theta = 30^\circ\) on both sides
\( \sin 30^\circ = \sqrt{\frac{1 - \cos 60^\circ}{2}} = \sqrt{\frac{1 - \frac{1}{2}}{2}} \) [\( \because \cos 60^\circ = \frac{1}{2} \) (Given)]
\( = \sqrt{\frac{1/2}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
Hence, \(\sin 30^\circ = \frac{1}{2}\).

SOLVED EXAMPLES

Question. Prove that \( \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta} \)
Answer: L.H.S. \( = \frac{\cos \theta}{1 + \sin \theta} = \frac{\cos \theta(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} \) [Multiplying numerator and denominator by \( (1 - \sin \theta) \)]
\( = \frac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta} = \frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta} \)
\( = \frac{1 - \sin \theta}{\cos \theta} = \text{R.H.S.} \)

Question. Prove that \( \tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta \)
Answer: L.H.S. \( = \tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \theta}{1} \)
\( = \frac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta(1 - \cos^2 \theta)}{\cos^2 \theta} \)
\( = \frac{\sin^2 \theta}{\cos^2 \theta} (1 - \cos^2 \theta) = \tan^2 \theta \sin^2 \theta = \text{R.H.S.} \) [Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( 1 - \cos^2 \theta = \sin^2 \theta \)]

Question. Prove that \( \tan^2 \theta + \cot^2 \theta + 2 = \sec^2 \theta \csc^2 \theta \)
Answer: L.H.S. \( = \tan^2 \theta + \cot^2 \theta + 2 = (\tan^2 \theta + 1) + (\cot^2 \theta + 1) \)
\( = \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \)
\( = \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \csc^2 \theta \sec^2 \theta = \text{R.H.S.} \)

Question. Prove that \( \frac{\sec \theta - 1}{\sec \theta + 1} = \frac{1 - \cos \theta}{1 + \cos \theta} = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 \)
Answer: L.H.S. \( = \frac{\sec \theta - 1}{\sec \theta + 1} = \frac{\frac{1}{\cos \theta} - 1}{\frac{1}{\cos \theta} + 1} = \frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{1 + \cos \theta}{\cos \theta}} = \frac{1 - \cos \theta}{1 + \cos \theta} = \text{R.H.S.} \)
Now, \( \frac{1 - \cos \theta}{1 + \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{(1 + \cos \theta)^2} = \frac{1 - \cos^2 \theta}{(1 + \cos \theta)^2} \)
\( = \frac{\sin^2 \theta}{(1 + \cos \theta)^2} = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 = \text{R.H.S.} \) Proved.

Question. Prove that \( \sec^2 \theta + \csc^2 \theta = \sec^2 \theta \csc^2 \theta \)
Answer: L.H.S. \( = \sec^2 \theta + \csc^2 \theta \)
\( = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \cdot \sin^2 \theta} = \frac{1}{\cos^2 \theta \cdot \sin^2 \theta} \)
\( = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta} = \sec^2 \theta \cdot \csc^2 \theta = \text{R.H.S.} \)

Question. Prove that \( \cot \theta - \tan \theta = \frac{2 \cos^2 \theta - 1}{\sin \theta \cos \theta} \)
Answer: L.H.S. \( = \cot \theta - \tan \theta \)
\( = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} \)
\( = \frac{\cos^2 \theta - (1 - \cos^2 \theta)}{\sin \theta \cos \theta} \) [Since \( \sin^2 \theta = 1 - \cos^2 \theta \)]
\( = \frac{\cos^2 \theta - 1 + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{2 \cos^2 \theta - 1}{\sin \theta \cos \theta} = \text{R.H.S.} \)

Question. Prove the identity: \( \frac{\cos^2 \theta}{\sin \theta} + \sin \theta = \csc \theta \)
Answer: L.H.S. \( = \frac{\cos^2 \theta}{\sin \theta} + \sin \theta = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta} \)
\( = \frac{1}{\sin \theta} = \csc \theta = \text{R.H.S.} \) [Since \( \sin^2 \theta + \cos^2 \theta = 1 \)]

Question. Prove the identity: \( \frac{1 - \tan^2 \theta}{\cot^2 \theta - 1} = \tan^2 \theta \)
Answer: L.H.S. \( = \frac{1 - \tan^2 \theta}{\cot^2 \theta - 1} = \frac{1 - \tan^2 \theta}{\frac{1}{\tan^2 \theta} - 1} = \frac{1 - \tan^2 \theta}{\frac{1 - \tan^2 \theta}{\tan^2 \theta}} \)
\( = \frac{(1 - \tan^2 \theta) \tan^2 \theta}{1 - \tan^2 \theta} = \tan^2 \theta = \text{R.H.S.} \)

Question. Prove the identity: \( (\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta} \)
Answer: L.H.S. \( = (\sec \theta - \tan \theta)^2 \)
\( = \sec^2 \theta + \tan^2 \theta - 2 \sec \theta \tan \theta = \frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{2 \sin \theta}{\cos \theta \cdot \cos \theta} \)
\( = \frac{1 + \sin^2 \theta - 2 \sin \theta}{\cos^2 \theta} = \frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta} \) [Since \( \cos^2 \theta = 1 - \sin^2 \theta \)]
\( = \frac{(1 - \sin \theta)(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{1 - \sin \theta}{1 + \sin \theta} = \text{R.H.S.} \) Proved.

Question. Show that \( (1 + \tan A \tan B)^2 + (\tan A - \tan B)^2 = \sec^2 A \sec^2 B \)
Answer: L.H.S. \( = (1 + \tan A \tan B)^2 + (\tan A - \tan B)^2 \)
\( = 1 + \tan^2 A \tan^2 B + 2 \tan A \tan B + \tan^2 A + \tan^2 B - 2 \tan A \tan B \)
\( = 1 + \tan^2 A + \tan^2 A \tan^2 B + \tan^2 B = (1 + \tan^2 A) + \tan^2 B (\tan^2 A + 1) \)
\( = \sec^2 A + \tan^2 B \sec^2 A = \sec^2 A (1 + \tan^2 B) = \sec^2 A \sec^2 B = \text{R.H.S.} \)

Question. Show that \( (1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) = 1 \)
Answer: L.H.S. \( = (1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) = (1 + \tan^2 \theta)(1 - \sin^2 \theta) \)
\( = (\sec^2 \theta)(\cos^2 \theta) = \left( \frac{1}{\cos^2 \theta} \right) (\cos^2 \theta) = 1 = \text{R.H.S.} \)

Question. Show that \( (\sec \theta + \cos \theta)(\sec \theta - \cos \theta) = \tan^2 \theta + \sin^2 \theta \)
Answer: L.H.S. \( = (\sec \theta + \cos \theta)(\sec \theta - \cos \theta) = \sec^2 \theta - \cos^2 \theta \)
\( = (\tan^2 \theta + 1) - \cos^2 \theta = \tan^2 \theta + (1 - \cos^2 \theta) = \tan^2 \theta + \sin^2 \theta = \text{R.H.S.} \)

Question. Show that \( \tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)
Answer: R.H.S. \( = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} = \frac{\sin^2 A}{\cos^2 A \cos^2 B} - \frac{\sin^2 B}{\cos^2 A \cos^2 B} \)
\( = \tan^2 A \sec^2 B - \tan^2 B \sec^2 A = \tan^2 A(1 + \tan^2 B) - \tan^2 B(1 + \tan^2 A) \)
\( = \tan^2 A + \tan^2 A \tan^2 B - \tan^2 B - \tan^2 B \tan^2 A = \tan^2 A - \tan^2 B = \text{L.H.S.} \) Proved.

Question. Prove that \( (1 + \cot^2 \theta)(1 - \cos \theta)(1 + \cos \theta) = 1 \)
Answer: L.H.S. \( = (1 + \cot^2 \theta)(1 - \cos \theta)(1 + \cos \theta) = (1 + \cot^2 \theta)(1 - \cos^2 \theta) \)
\( = (\csc^2 \theta) \sin^2 \theta = \frac{1}{\sin^2 \theta} \cdot \sin^2 \theta = 1 = \text{R.H.S.} \)

Question. Prove that \( \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = 1 - 2 \sec \theta \tan \theta + 2 \tan^2 \theta \)
Answer: L.H.S. \( = \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} \times \frac{\sec \theta - \tan \theta}{\sec \theta - \tan \theta} \)
\( = \frac{(\sec \theta - \tan \theta)^2}{\sec^2 \theta - \tan^2 \theta} = \frac{\sec^2 \theta + \tan^2 \theta - 2 \sec \theta \tan \theta}{1} \) [Since \( \sec^2 \theta - \tan^2 \theta = 1 \)]
\( = (1 + \tan^2 \theta) + \tan^2 \theta - 2 \sec \theta \tan \theta = 1 - 2 \sec \theta \tan \theta + 2 \tan^2 \theta = \text{R.H.S.} \)

Question. Prove that \( \frac{\tan \theta - \cot \theta}{\sin \theta \cos \theta} = \tan^2 \theta - \cot^2 \theta \)
Answer: L.H.S. \( = \frac{\tan \theta - \cot \theta}{\sin \theta \cos \theta} = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta}}{\sin \theta \cos \theta} = \frac{\frac{\sin^2 \theta - \cos^2 \theta}{\cos \theta \sin \theta}}{\sin \theta \cos \theta} \)
\( = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta} = \sec^2 \theta - \csc^2 \theta \)
\( = (1 + \tan^2 \theta) - (1 + \cot^2 \theta) = \tan^2 \theta - \cot^2 \theta = \text{R.H.S.} \)

Question. Prove that \( \frac{\sin \theta}{1 - \cos \theta} + \frac{\tan \theta}{1 + \cos \theta} = \sec \theta \csc \theta + \cot \theta \)
Answer: L.H.S. \( = \frac{\sin \theta}{1 - \cos \theta} + \frac{\tan \theta}{1 + \cos \theta} = \frac{\sin \theta(1 + \cos \theta) + \tan \theta(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \)
\( = \frac{\sin \theta + \sin \theta \cos \theta + \tan \theta - \tan \theta \cos \theta}{1 - \cos^2 \theta} = \frac{\sin \theta + \sin \theta \cos \theta + \tan \theta - \left( \frac{\sin \theta}{\cos \theta} \right) \cos \theta}{\sin^2 \theta} \)
\( = \frac{\sin \theta + \sin \theta \cos \theta + \tan \theta - \sin \theta}{\sin^2 \theta} = \frac{\sin \theta \cos \theta + \tan \theta}{\sin^2 \theta} \)
\( = \frac{\sin \theta \cos \theta}{\sin^2 \theta} + \frac{\tan \theta}{\sin^2 \theta} = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta \sin^2 \theta} = \cot \theta + \frac{1}{\cos \theta \sin \theta} \)
\( = \cot \theta + \sec \theta \csc \theta = \text{R.H.S.} \) Proved.

Question. Solve the equation: \( \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4 \)
Answer: \( \frac{\cos \theta(1 + \sin \theta) + \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = 4 \)
\( \implies \frac{\cos \theta + \cos \theta \sin \theta + \cos \theta - \cos \theta \sin \theta}{1 - \sin^2 \theta} = 4 \)
\( \implies \frac{2 \cos \theta}{\cos^2 \theta} = 4 \)
\( \implies \frac{2}{\cos \theta} = 4 \)
\( \implies \cos \theta = \frac{2}{4} = \frac{1}{2} \)
\( \implies \cos \theta = \cos 60^\circ \)
\( \implies \theta = 60^\circ \) Ans.

Question. Solve the equation: \( \frac{\cos^2 \theta - 3 \cos \theta + 2}{\sin^2 \theta} = 1 \), \( \sin \theta \neq 0 \)
Answer: \( \cos^2 \theta - 3 \cos \theta + 2 = \sin^2 \theta \)
\( \implies \cos^2 \theta - 3 \cos \theta + 2 = 1 - \cos^2 \theta \)
\( \implies 2 \cos^2 \theta - 3 \cos \theta + 1 = 0 \)
Using quadratic formula, \( \cos \theta = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \)
\( \implies \cos \theta = 1 \) or \( \frac{1}{2} \)
So, either \( \cos \theta = 1 \implies \theta = 0^\circ \) (Not possible as \( \sin \theta \neq 0 \))
or \( \cos \theta = \frac{1}{2} \implies \theta = 60^\circ \). Hence, \( \theta = 60^\circ \) Ans.

Question. Solve the equation: \( \frac{\cos \theta}{\csc \theta + 1} + \frac{\cos \theta}{\csc \theta - 1} = 2 \)
Answer: \( \frac{\cos \theta}{\frac{1}{\sin \theta} + 1} + \frac{\cos \theta}{\frac{1}{\sin \theta} - 1} = 2 \)
\( \implies \frac{\sin \theta \cos \theta}{1 + \sin \theta} + \frac{\sin \theta \cos \theta}{1 - \sin \theta} = 2 \)
\( \implies \frac{\sin \theta \cos \theta (1 - \sin \theta) + \sin \theta \cos \theta (1 + \sin \theta)}{1 - \sin^2 \theta} = 2 \)
\( \implies \frac{\sin \theta \cos \theta [1 - \sin \theta + 1 + \sin \theta]}{\cos^2 \theta} = 2 \)
\( \implies \frac{2 \sin \theta \cos \theta}{\cos^2 \theta} = 2 \)
\( \implies \frac{\sin \theta}{\cos \theta} = 1 \)
\( \implies \tan \theta = 1 \)
\( \implies \theta = 45^\circ \) Ans.

Question. Prove the identity: \( (1 + \tan^2 A) + \left( 1 + \frac{1}{\tan^2 A} \right) = \frac{1}{\sin^2 A - \sin^4 A} \)
Answer: L.H.S. \( = \sec^2 A + (1 + \cot^2 A) = \sec^2 A + \csc^2 A \)
\( = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A} = \frac{1}{\sin^2 A(1 - \sin^2 A)} \)
\( = \frac{1}{\sin^2 A - \sin^4 A} = \text{R.H.S.} \) Proved.

Question. Prove the identity: \( \sec^6 \theta = \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta + 1 \)
Answer: L.H.S. \( = \sec^6 \theta = (\sec^2 \theta)^3 = (1 + \tan^2 \theta)^3 \)
\( = 1 + \tan^6 \theta + 3(1)(\tan^2 \theta)(1 + \tan^2 \theta) \)
\( = 1 + \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta = \text{R.H.S.} \)

Question. Prove the identity: \( \frac{(1 + \tan^2 \theta) \cot \theta}{\csc^2 \theta} = \tan \theta \)
Answer: L.H.S. \( = \frac{\sec^2 \theta \cdot \cot \theta}{\csc^2 \theta} = \frac{\frac{1}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin^2 \theta}} \)
\( = \frac{1}{\cos \theta \sin \theta} \cdot \sin^2 \theta = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{R.H.S.} \)

Question. Prove the identity: \( \frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A} = \sec^2 A \csc^2 A - 2 \)
Answer: L.H.S. \( = \frac{\sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A} = \frac{(\sin^2 A)^2 + (\cos^2 A)^2 + 2 \sin^2 A \cos^2 A - 2 \sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \)
\( = \frac{(\sin^2 A + \cos^2 A)^2 - 2 \sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} = \frac{1 - 2 \sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \)
\( = \frac{1}{\sin^2 A \cos^2 A} - 2 = \sec^2 A \csc^2 A - 2 = \text{R.H.S.} \)

Question. Prove the identity: \( \cot^4 A - 1 = \csc^4 A - 2 \csc^2 A \)
Answer: L.H.S. \( = \cot^4 A - 1 = (\cot^2 A)^2 - 1 = (\csc^2 A - 1)^2 - 1 \)
\( = \csc^4 A + 1 - 2 \csc^2 A - 1 = \csc^4 A - 2 \csc^2 A = \text{R.H.S.} \)

Question. Prove the identity: \( \sin^8 \theta - \cos^8 \theta = (\sin^2 \theta - \cos^2 \theta)(1 - 2 \sin^2 \theta \cos^2 \theta) \)
Answer: L.H.S. \( = (sin^4 \theta)^2 - (\cos^4 \theta)^2 = (\sin^4 \theta + \cos^4 \theta)(\sin^4 \theta - \cos^4 \theta) \)
\( = [(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta] [(\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta)] \)
\( = (1 - 2 \sin^2 \theta \cos^2 \theta) (1) (\sin^2 \theta - \cos^2 \theta) = \text{R.H.S.} \)

Question. Prove the identity: \( \sec^4 A(1 - \sin^4 A) - 2 \tan^2 A = 1 \)
Answer: L.H.S. \( = \sec^4 A - \sec^4 A \sin^4 A - 2 \tan^2 A = \sec^4 A - \tan^4 A - 2 \tan^2 A \)
\( = (\sec^2 A)^2 - [(\tan^2 A)^2 + 2 \tan^2 A + 1] + 1 = (\sec^2 A)^2 - (1 + \tan^2 A)^2 + 1 \)
\( = (\sec^2 A)^2 - (\sec^2 A)^2 + 1 = 1 = \text{R.H.S.} \) Proved.

Question. Prove the identity: \( (1 + \tan^2 \theta) \cos^2 \theta = 1 \)
Answer: L.H.S. \( = (1 + \frac{\sin^2 \theta}{\cos^2 \theta}) \cos^2 \theta = \left( \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} \right) \cos^2 \theta = 1 = \text{R.H.S.} \)

Question. Prove the identity: \( \sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta \)
Answer: L.H.S. \( = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = (\sin^2 \theta - \cos^2 \theta)(1) = \text{R.H.S.} \)

Question. Prove the identity: \( \frac{\csc^2 \theta - 1}{\csc^2 \theta} = \cos^2 \theta \)
Answer: L.H.S. \( = \frac{\frac{1}{\sin^2 \theta} - 1}{\frac{1}{\sin^2 \theta}} = \frac{1 - \sin^2 \theta}{\sin^2 \theta} \cdot \frac{\sin^2 \theta}{1} = 1 - \sin^2 \theta = \cos^2 \theta = \text{R.H.S.} \)

Question. Prove the identity: \( (\cos^2 \theta - 1)(\cot^2 \theta + 1) + 1 = 0 \)
Answer: L.H.S. \( = (\cos^2 \theta - 1) \csc^2 \theta + 1 = - (1 - \cos^2 \theta) \csc^2 \theta + 1 \)
\( = - \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} + 1 = -1 + 1 = 0 = \text{R.H.S.} \) Proved.