CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 07

Question. Show that \( \frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B \)
Answer: L.H.S. \( = \frac{\frac{\cos A}{\sin A} + \frac{\sin B}{\cos B}}{\frac{\cos B}{\sin B} + \frac{\sin A}{\cos A}} = \frac{\frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B}}{\frac{\sin A \cos B + \cos A \sin B}{\cos A \sin B}} \)
\( = \frac{\cos A \sin B}{\sin A \cos B} = \cot A \tan B = \text{R.H.S.} \)

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

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

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

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

Question. Prove that \( \frac{\sin A - \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A + \sin B} = 0 \)
Answer: L.H.S. \( = \frac{\sin^2 A - \sin^2 B + \cos^2 A - \cos^2 B}{(\cos A + \cos B)(\sin A + \sin B)} = \frac{(\sin^2 A + \cos^2 A) - (\sin^2 B + \cos^2 B)}{(\cos A + \cos B)(\sin A + \sin B)} \)
\( = \frac{1 - 1}{(\cos A + \cos B)(\sin A + \sin B)} = 0 = \text{R.H.S.} \)

Question. Prove that \( \sec^4 \theta - \tan^4 \theta = 1 + 2 \tan^2 \theta \)
Answer: L.H.S. \( = (\sec^2 \theta + \tan^2 \theta)(\sec^2 \theta - \tan^2 \theta) = (\sec^2 \theta + \tan^2 \theta)(1) \)
\( = (1 + \tan^2 \theta) + \tan^2 \theta = 1 + 2 \tan^2 \theta = \text{R.H.S.} \)

Question. Prove that \( \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + \cos^2 A \cos^2 B + \sin^2 A \sin^2 B = 1 \)
Answer: L.H.S. \( = \sin^2 A (\cos^2 B + \sin^2 B) + \cos^2 A (\sin^2 B + \cos^2 B) \)
\( = \sin^2 A (1) + \cos^2 A (1) = 1 = \text{R.H.S.} \)

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

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

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

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

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

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

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

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

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

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

Question. Show that: \( \frac{\sin \theta}{\cot \theta + \csc \theta} = 2 + \frac{\sin \theta}{\cot \theta - \csc \theta} \)
Answer: L.H.S. \( = \frac{\sin \theta}{\frac{\cos \theta + 1}{\sin \theta}} = \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \).
Now, R.H.S. \( = 2 + \frac{\sin \theta}{\frac{\cos \theta - 1}{\sin \theta}} = 2 + \frac{\sin^2 \theta}{-(1 - \cos \theta)} \)
\( = 2 - (1 + \cos \theta) = 1 - \cos \theta \). Clearly, L.H.S. = R.H.S. Proved.

Question. Prove that \( \frac{(1 + \cot A + \tan A)(\sin A - \cos A)}{\sec^3 A - \csc^3 A} = \sin^2 A \cos^2 A \)
Answer: L.H.S. \( = \frac{\left( \frac{\sin A \cos A + \cos^2 A + \sin^2 A}{\sin A \cos A} \right) (\sin A - \cos A)}{(\sec A - \csc A)(\sec^2 A + \sec A \csc A + \csc^2 A)} \)
\( = \frac{\frac{(\sin A \cos A + 1)(\sin A - \cos A)}{\sin A \cos A}}{(\sec A - \csc A) \left( \frac{\sin^2 A + \sin A \cos A + \cos^2 A}{\sin^2 A \cos^2 A} \right)} \)
\( = \frac{(\sin A \cos A + 1)(\sec A - \csc A) \cdot \sin^2 A \cos^2 A}{(\sec A - \csc A)(1 + \sin A \cos A)} = \sin^2 A \cos^2 A = \text{R.H.S.} \) Proved.

SOLVED EXAMPLES

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

Question. Prove that \( \left( \frac{1}{\sec^2 \theta - \cos^2 \theta} + \frac{1}{\csc^2 \theta - \sin^2 \theta} \right) \sin^2 \theta \cos^2 \theta = \frac{1 - \sin^2 \theta \cos^2 \theta}{2 + \sin^2 \theta \cos^2 \theta} \)
Answer: L.H.S. \( = \left( \frac{1}{\sec^2 \theta - \cos^2 \theta} + \frac{1}{\csc^2 \theta - \sin^2 \theta} \right) \sin^2 \theta \cos^2 \theta \)
\( = \left( \frac{1}{\frac{1}{\cos^2 \theta} - \cos^2 \theta} + \frac{1}{\frac{1}{\sin^2 \theta} - \sin^2 \theta} \right) \sin^2 \theta \cos^2 \theta \)
\( = \left( \frac{\cos^2 \theta}{1 - \cos^4 \theta} + \frac{\sin^2 \theta}{1 - \sin^4 \theta} \right) \sin^2 \theta \cos^2 \theta \)
\( = \left( \frac{\cos^2 \theta}{(1 + \cos^2 \theta)(1 - \cos^2 \theta)} + \frac{\sin^2 \theta}{(1 - \sin^2 \theta)(1 + \sin^2 \theta)} \right) \sin^2 \theta \cos^2 \theta \)
\( = \left[ \frac{\cos^2 \theta}{(1 + \cos^2 \theta) \sin^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta (1 + \sin^2 \theta)} \right] \sin^2 \theta \cos^2 \theta \)
\( = \frac{\cos^4 \theta}{1 + \cos^2 \theta} + \frac{\sin^4 \theta}{1 + \sin^2 \theta} = \frac{[\cos^4 \theta(1 + \sin^2 \theta) + \sin^4 \theta(1 + \cos^2 \theta)]}{(1 + \cos^2 \theta)(1 + \sin^2 \theta)} \)
\( = \frac{\cos^4 \theta + \sin^2 \theta \cos^4 \theta + \sin^4 \theta + \sin^4 \theta \cos^2 \theta}{(1 + \cos^2 \theta)(1 + \sin^2 \theta)} \)
\( = \frac{\sin^4 \theta + \cos^4 \theta + \sin^2 \theta \cos^2 \theta (\cos^2 \theta + \sin^2 \theta)}{(1 + \cos^2 \theta)(1 + \sin^2 \theta)} \)
\( = \frac{(\sin^2 \theta)^2 + (\cos^2 \theta)^2 + 2 \sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta}{1 + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \cos^2 \theta} \)
\( = \frac{(\sin^2 \theta + \cos^2 \theta)^2 - \sin^2 \theta \cos^2 \theta}{1 + (\sin^2 \theta + \cos^2 \theta) + \sin^2 \theta \cos^2 \theta} \)
\( = \frac{1 - \sin^2 \theta \cos^2 \theta}{1 + 1 + \sin^2 \theta \cos^2 \theta} \)
\( = \frac{1 - \sin^2 \theta \cos^2 \theta}{2 + \sin^2 \theta \cos^2 \theta} = \text{R.H.S.} \)
Proved.

Question. Express the trigonometric ratios \( \sin A \), \( \sec A \), and \( \tan A \) in terms of \( \cot A \).
Answer: (i) We know that \( \csc^2 A = 1 + \cot^2 A \)
\( \implies \frac{1}{\sin^2 A} = 1 + \cot^2 A \)
\( \implies \sin^2 A = \frac{1}{1 + \cot^2 A} \)
\( \implies \sin A = \frac{1}{\sqrt{1 + \cot^2 A}} \)
(ii) Also, we know that \( \sec^2 A = 1 + \tan^2 A \)
\( \implies \sec^2 A = 1 + \frac{1}{\cot^2 A} \)
\( \implies \sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A} \)
\( \implies \sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A} \)
(iii) Also we know that, \( \tan A = \frac{1}{\cot A} \)

Question. Evaluate: (i) \( \frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ} \) (ii) \( \sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ \)
Answer: (i) \( \frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ} = \frac{\sin^2 (90^\circ - 27^\circ) + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 (90^\circ - 17^\circ)} = \frac{\cos^2 27^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \sin^2 17^\circ} \) [Since \( \sin^2 \theta + \cos^2 \theta = 1 \)]
\( = \frac{1}{1} = 1 \)
(ii) \( \sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ \)
\( = \sin 25^\circ \cos (90^\circ - 25^\circ) + \cos 25^\circ \sin (90^\circ - 25^\circ) \)
\( = \sin 25^\circ \sin 25^\circ + \cos 25^\circ \cos 25^\circ \)
\( = \sin^2 25^\circ + \cos^2 25^\circ = 1 \) [Since \( \sin^2 \theta + \cos^2 \theta = 1 \)]

Question. \( 9 \sec^2 A - 9 \tan^2 A = \)
(a) 1
(b) 9
(c) 8
(d) 0
Answer: (b) 9
Justification: \( 9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A) = 9 \times 1 = 9 \)

Question. \( (1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta) = \)
(a) 0
(b) 1
(c) 2
(d) -1
Answer: (c) 2
Justification: \( (1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta})(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}) = \left( \frac{\cos \theta + \sin \theta + 1}{\cos \theta} \right) \left( \frac{\sin \theta + \cos \theta - 1}{\sin \theta} \right) \)
\( = \frac{(\cos \theta + \sin \theta)^2 - 1}{\sin \theta \cos \theta} = \frac{(\cos^2 \theta + \sin^2 \theta) + 2 \cos \theta \sin \theta - 1}{\sin \theta \cos \theta} = \frac{1 + 2 \cos \theta \sin \theta - 1}{\sin \theta \cos \theta} \)
\( = \frac{2 \cos \theta \sin \theta}{\sin \theta \cos \theta} = 2 \)

Question. \( (\sec A + \tan A)(1 - \sin A) = \)
(a) \( \sec A \)
(b) \( \sin A \)
(c) \( \csc A \)
(d) \( \cos A \)
Answer: (d) \( \cos A \)
Justification: \( (\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A) = \left( \frac{1 + \sin A}{\cos A} \right) (1 - \sin A) \)
\( = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \)

Question. \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \)
(a) \( \sec^2 A \)
(b) -1
(c) \( \cot^2 A \)
(d) \( \tan^2 A \)
Answer: (d) \( \tan^2 A \)
Justification: \( \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}} = \frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \times \frac{\tan^2 A}{1 + \tan^2 A} = \tan^2 A \)

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

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

Question. Prove the identity: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \sec \theta \csc \theta + 1 \)
Answer: L.H.S. \( = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta - \sin \theta} \)
\( = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
\( = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{(\sin \theta - \cos \theta) \sin \theta \cos \theta} \) [Since \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)]
\( = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \sec \theta \csc \theta + 1 = \text{R.H.S.} \)
Proved.

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

Question. Prove the identity: \( \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A \)
Answer: Dividing numerator and denominator by \( \sin A \), we get
L.H.S. \( = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A} = \frac{\cot A + \csc A - 1}{1 - \csc A + \cot A} \)
\( = \frac{\cot A + \csc A - (\csc^2 A - \cot^2 A)}{1 - \csc A + \cot A} \) [Since \( \csc^2 A = 1 + \cot^2 A \)]
\( = \frac{(\cot A + \csc A) - (\csc A + \cot A)(\csc A - \cot A)}{1 - \csc A + \cot A} \)
\( = \frac{(\cot A + \csc A)(1 - \csc A + \cot A)}{1 - \csc A + \cot A} = \cot A + \csc A = \text{R.H.S.} \)
Proved.

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

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

Question. Prove the identity: \( (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)
Answer: L.H.S. \( = \sin^2 A + \csc^2 A + 2 \sin A \csc A + \cos^2 A + \sec^2 A + 2 \cos A \sec A \)
\( = (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2 \frac{\sin A}{\sin A} + 2 \frac{\cos A}{\cos A} \)
\( = 1 + \csc^2 A + \sec^2 A + 2 + 2 = (1 + \cot^2 A) + (1 + \tan^2 A) + 5 \)
\( = 7 + \cot^2 A + \tan^2 A = \text{R.H.S.} \)
Proved.

Question. Prove the identity: \( (\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A} \)
Answer: L.H.S. \( = \csc A \sec A - \csc A \cos A - \sin A \sec A + \sin A \cos A \)
\( = \frac{1}{\sin A \cos A} - \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} + \sin A \cos A \)
\( = \frac{1 - \cos^2 A - \sin^2 A + \sin^2 A \cos^2 A}{\sin A \cos A} = \frac{1 - (\cos^2 A + \sin^2 A) + \sin^2 A \cos^2 A}{\sin A \cos A} \)
\( = \frac{1 - 1 + \sin^2 A \cos^2 A}{\sin A \cos A} = \frac{\sin^2 A \cos^2 A}{\sin A \cos A} = \sin A \cos A \) ... (1)
R.H.S. \( = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{\sin A \cos A}{1} = \sin A \cos A \) ... (2)
From (1) and (2), we have L.H.S. = R.H.S. Proved.

Question. Prove the identity: \( \frac{1 + \tan^2 \theta}{1 + \cot^2 \theta} = \left( \frac{1 - \tan \theta}{1 - \cot \theta} \right)^2 = \tan^2 \theta \)
Answer: L.H.S. \( = \frac{\sec^2 \theta}{\csc^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \) ... (1)
Now, R.H.S. \( = \left( \frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}} \right)^2 = \left( \frac{1 - \tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} \right)^2 = \left( \frac{1 - \tan \theta}{-(1 - \tan \theta)} \cdot \tan \theta \right)^2 = (-\tan \theta)^2 = \tan^2 \theta \) ... (2)
From (1) and (2), clearly, L.H.S. = R.H.S. Proved.