CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set 04

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Short Answer Type Questions

Question. Prove that: \( (\sqrt{3} + 1)(3 - \cot 30^\circ) = \tan^3 60^\circ - 2 \sin 60^\circ \)
Answer: Sol. L.H.S. \( = (\sqrt{3} + 1)(3 - \cot 30^\circ) = (\sqrt{3} + 1)(3 - \sqrt{3}) \)
...[ \( \because \cot 30^\circ = \sqrt{3} \)
\( = 3\sqrt{3} - \sqrt{3} \cdot \sqrt{3} + 3 - \sqrt{3} \)
\( = 3\sqrt{3} - 3 + 3 - \sqrt{3} = 2\sqrt{3} \)
Also, R.H.S. \( = \tan^3 60^\circ - 2 \sin 60^\circ \)
\( = (\sqrt{3})^3 - 2 \frac{\sqrt{3}}{2} = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \)
\( \therefore \) L.H.S. = R.H.S. ...(Hence proved)

Question. If \( \sqrt{3} \tan \theta = 1 \), then find the value of \( \sin^2 \theta - \cos^2 \theta \).
Answer: Sol. Given: \( \sqrt{3} \tan \theta = 1 \)
\( \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \tan \theta = \tan 30^\circ \)
\( \Rightarrow \theta = 30^\circ \)
\( \therefore \sin^2 \theta - \cos^2 \theta = \sin^2 30^\circ - \cos^2 30^\circ \)
\( = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{3}{4} = \frac{1-3}{4} = \frac{-2}{4} = -\frac{1}{2} \)

Question. Simplify \( (1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) \).
Answer: Sol. \( (1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) \)
\( = (\sec^2 \theta)(1 - \sin^2 \theta) \) ...[\( \because 1 + \tan^2 \theta = \sec^2 \theta \); \( (a - b)(a + b) = a^2 - b^2 \)]
\( = \sec^2 \theta \cos^2 \theta \) ...[\( \because \sin^2 \theta + \cos^2 \theta = 1 \)]
\( = \frac{1}{\cos^2 \theta} \cdot \cos^2 \theta = 1 \)

Question. Show that \( \tan^4 \theta + \tan^2 \theta = \sec^4 \theta - \sec^2 \theta \).
Answer: Sol. L.H.S. \( \tan^4 \theta + \tan^2 \theta = \tan^2 \theta (\tan^2 \theta + 1) \)
\( = \tan^2 \theta \cdot \sec^2 \theta \) ...[\( \because \sec^2 \theta = \tan^2 \theta + 1 \)]
\( = (\sec^2 \theta - 1) \sec^2 \theta \) ...[\( \because \tan^2 \theta = \sec^2 \theta - 1 \)]
\( = \sec^4 \theta - \sec^2 \theta = \text{R.H.S.} \) ...(Hence proved)

Long Answer Type Questions

Question. If \( \text{cosec } \theta + \cot \theta = P \), then prove that \( \cos \theta = \frac{P^2 - 1}{P^2 + 1} \).
Answer: Sol. R.H.S \( = \frac{P^2 - 1}{P^2 + 1} \)
\( = \frac{(\text{cosec } \theta + \cot \theta)^2 - 1}{(\text{cosec } \theta + \cot \theta)^2 + 1} \)
\( = \frac{\text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta + \cot^2 \theta - 1}{\text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta + \cot^2 \theta + 1} \)
\( = \frac{\cot^2 \theta + \cot^2 \theta + 2 \text{cosec } \theta \cot \theta}{\text{cosec}^2 \theta + \text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta} \) ...[\( \text{cosec}^2 \theta - 1 = \cot^2 \theta \)]
\( = \frac{2 \cot^2 \theta + 2 \text{cosec } \theta \cot \theta}{2 \text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta} \)
\( = \frac{2 \cot \theta (\cot \theta + \text{cosec } \theta)}{2 \text{cosec } \theta (\text{cosec } \theta + \cot \theta)} = \frac{\cot \theta}{\text{cosec } \theta} \)
\( = \frac{\cos \theta}{\sin \theta} \times \frac{\sin \theta}{1} = \cos \theta = \text{L.H.S.} \)

Question. Prove that \( \sqrt{\sec^2 \theta + \text{cosec}^2 \theta} = \tan \theta + \cot \theta \).
Answer: Sol. L.H.S. \( = \sqrt{\sec^2 \theta + \text{cosec}^2 \theta} \)
\( = \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}} \) ...[\( \sec^2 \theta = 1/\cos^2 \theta \); \( \text{cosec}^2 \theta = 1/\sin^2 \theta \)]
\( = \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}} \) ...[\( \because \sin^2 \theta + \cos^2 \theta = 1 \)]
\( = \frac{1}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \) ...[\( \because 1 = \sin^2 \theta + \cos^2 \theta \)]
\( = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \tan \theta + \cot \theta = \text{R.H.S.} \) ...(Hence proved)

Question. If \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \), then prove that \( \tan \theta = 1 \) or \( \frac{1}{2} \).
Answer: Sol. \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \) ...[Given
Dividing both sides by \( \sin^2 \theta \), we get
\( \frac{1}{\sin^2 \theta} + 1 = 3 \cdot \frac{\cos \theta}{\sin \theta} \)
\( \Rightarrow \text{cosec}^2 \theta + 1 = 3 \cdot \cot \theta \)
\( \Rightarrow 1 + \cot^2 \theta + 1 = 3 \cot \theta \) ...[\( \because \text{cosec}^2 \theta - \cot^2 \theta = 1 \)]
\( \Rightarrow \cot^2 \theta - 3 \cot \theta + 2 = 0 \)
\( \Rightarrow \cot^2 \theta - 2 \cot \theta - \cot \theta + 2 = 0 \)
\( \Rightarrow \cot \theta(\cot \theta - 2) - 1(\cot \theta - 2) = 0 \)
\( \Rightarrow (\cot \theta - 2)(\cot \theta - 1) = 0 \)
\( \cot \theta - 1 = 0 \text{ or } \cot \theta - 2 = 0 \)
\( \Rightarrow \cot \theta = 1 \text{ or } \cot \theta = 2 \)
\( \therefore \tan \theta = 1 \text{ or } \frac{1}{2} \) ...[\( \because \tan \theta = \frac{1}{\cot \theta} \)]
...(Hence proved)

Question. If \( \sin \theta + 2 \cos \theta = 1 \), then prove that \( 2 \sin \theta - \cos \theta = 2 \).
Answer: Sol. Given: \( \sin \theta + 2 \cos \theta = 1 \)
On squaring both sides, we get
\( (\sin \theta + 2 \cos \theta)^2 = 1^2 \)
\( \Rightarrow \sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 1 \)
\( \Rightarrow (1 - \cos^2 \theta) + 4(1 - \sin^2 \theta) + 4 \sin \theta \cos \theta = 1 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]
\( \Rightarrow 1 - \cos^2 \theta + 4 - 4 \sin^2 \theta + 4 \sin \theta \cos \theta = 1 \)
\( \Rightarrow - \cos^2 \theta - 4 \sin^2 \theta + 4 \sin \theta \cos \theta = -4 \)
\( \Rightarrow 4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = 4 \)
\( \Rightarrow (2 \sin \theta - \cos \theta)^2 = 4 \) ...[\( a^2 + b^2 - 2ab = (a - b)^2 \)]
\( \therefore 2 \sin \theta - \cos \theta = 2 \) ...(Hence proved)

Question. If \( \tan \theta + \sec \theta = l \), then prove that \( \sec \theta = \frac{l^2 + 1}{2l} \).
Answer: Sol. Given: \( \tan \theta + \sec \theta = l \) ...(i)
Multiplying the numerator and denominator of L.H.S. of above equation by \( (\sec \theta - \tan \theta) \), we get,
\( (\tan \theta + \sec \theta) \times \frac{\sec \theta - \tan \theta}{\sec \theta - \tan \theta} = l \)
\( \Rightarrow \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = l \) ...[\( (a+b)(a-b) = a^2 - b^2 \)]
\( \Rightarrow \frac{1}{\sec \theta - \tan \theta} = l \) ...[\( \sec^2 \theta - \tan^2 \theta = 1 \)]
\( \Rightarrow \sec \theta - \tan \theta = \frac{1}{l} \) ...(ii)
Adding equations (i) and (ii), we get
\( 2 \sec \theta = l + \frac{1}{l} \)
\( \Rightarrow \sec \theta = \frac{l^2 + 1}{2l} \) ...(Hence Proved)

Question. If \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \text{cosec } \theta = q \), then prove that \( q(p^2 - 1) = 2p \).
Answer: Sol. Given: \( \sin \theta + \cos \theta = p \) ...(i)
and \( \sec \theta + \text{cosec } \theta = q \)
\( \Rightarrow \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q \)
\( \Rightarrow \frac{\sin \theta + \cos \theta}{\sin \theta \cdot \cos \theta} = q \)
\( \Rightarrow \frac{p}{\sin \theta \cdot \cos \theta} = q \) ...[From (i)]
\( \Rightarrow \sin \theta \cdot \cos \theta = \frac{p}{q} \) ...(ii)
Squaring both sides of (i), we get
\( (\sin \theta + \cos \theta)^2 = p^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cdot \cos \theta = p^2 \) ...[\( (a+b)^2 = a^2 + 2ab + b^2 \)]
\( \Rightarrow 1 + 2 \sin \theta \cdot \cos \theta = p^2 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]
\( \Rightarrow 1 + 2 \cdot \frac{p}{q} = p^2 \) ...[From (ii)]
\( \Rightarrow \frac{q + 2p}{q} = p^2 \Rightarrow q + 2p = p^2q \)
\( \Rightarrow 2p = p^2q - q \)
\( \Rightarrow q(p^2 - 1) = 2p \) ...(Hence proved)

Question. If \( a \sin \theta + b \cos \theta = c \), then prove that \( a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} \).
Answer: Sol. Given: \( a \sin \theta + b \cos \theta = c \)
Squaring both sides of above equation, we get,
\( (a \sin \theta + b \cos \theta)^2 = c^2 \)
\( \Rightarrow a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = c^2 \)
...[\( (a+b)^2 = a^2 + b^2 + 2ab \)]
\( \Rightarrow a^2(1 - \cos^2 \theta) + b^2(1 - \sin^2 \theta) + 2ab \sin \theta \cos \theta = c^2 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]
\( \Rightarrow a^2 - a^2 \cos^2 \theta + b^2 - b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta = c^2 \)
\( \Rightarrow a^2 + b^2 - c^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta \)
\( \Rightarrow a^2 + b^2 - c^2 = (a \cos \theta - b \sin \theta)^2 \) ...[\( a^2 + b^2 - 2ab = (a - b)^2 \)]
\( \Rightarrow a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} \)
...(Hence proved)

Question. Prove that \( \frac{1 + \sec \theta - \tan \theta}{1 + \sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta} \).
Answer: Sol. L.H.S. \( = \frac{1 + \sec \theta - \tan \theta}{1 + \sec \theta + \tan \theta} \)
\( = \frac{(\sec^2 \theta - \tan^2 \theta) + \sec \theta - \tan \theta}{1 + \sec \theta + \tan \theta} \) ...[\( \sec^2 \theta - \tan^2 \theta = 1 \)]
\( = \frac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta)}{1 + \sec \theta + \tan \theta} \) ...[\( a^2 - b^2 = (a+b)(a-b) \)]
\( = \frac{(\sec \theta - \tan \theta)[(\sec \theta + \tan \theta) + 1]}{1 + \sec \theta + \tan \theta} \)
\( = \sec \theta - \tan \theta \)
\( = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \)
\( = \frac{1 - \sin \theta}{\cos \theta} = \text{R.H.S.} \) ...(Hence proved)

TRIGONOMETRIC RATIO (T.R.) OF SOME SPECIFIC ANGLES

The angles \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \) are angles for which we have values of T.R.

• \( \sin \theta \uparrow \) when \( \theta \uparrow, 0 \le \theta \le 90^\circ \)
• \( \cos \theta \downarrow \) when \( \theta \uparrow, 0 \le \theta \le 90^\circ \)
• \( \tan \theta, \cot \theta \) are not defined for \( \theta = 90^\circ \) & \( 0^\circ \) respectively.
• \( \csc \theta, \sec \theta \) are not defined when \( \theta = 0^\circ \) & \( 90^\circ \) respectively.
• \( \sin \theta = \cos \theta \) for only \( \theta = 45^\circ \)
• \( 180^\circ = \pi^c \)
• \( 30^\circ = \left( \frac{\pi}{6} \right)^c \); \( 45^\circ = \left( \frac{\pi}{4} \right)^c \); \( 60^\circ = \left( \frac{\pi}{3} \right)^c \); \( 90^\circ = \left( \frac{\pi}{2} \right)^c \)

EXAMPLES

Question. Evaluate each of the following in the simplest form :
(i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
(ii) \( \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
Answer: (i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \)
(ii) \( \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Evaluate the following expression :
(i) \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \)
(ii) \( 4\cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ \)
Answer: (i) \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \)
\( = \sqrt{3} \times (\sqrt{2})^2 + (2)^2 \times 1 = \sqrt{3} \times 2 + 4 = 4 + 2\sqrt{3} \)
(ii) \( 4\cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ \)
\( = 4(1)^2 - (2)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + 0 = 4 - 4 + \frac{3}{4} + 0 = \frac{3}{4} \)

Question. Show that :
(i) \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 30^\circ) = 6 \)
(ii) \( 2(\cos^4 60^\circ + \sin^4 30^\circ) - (\tan^2 60^\circ + \cot^2 45^\circ) + 3\sec^2 30^\circ = \frac{1}{4} \)
Answer: (i) \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 30^\circ) \)
\( = 2 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 + (\sqrt{3})^2 \right] - 6 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2 \right] \)
\( = 2 \left( \frac{1}{2} + 3 \right) - 6 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{7}{2} \right) - 6 \left( \frac{1}{6} \right) = 7 - 1 = 6 \)
(ii) \( 2(\cos^4 60^\circ + \sin^4 30^\circ) - (\tan^2 60^\circ + \cot^2 45^\circ) + 3\sec^2 30^\circ \)
\( = 2 \left[ \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 \right] - \left[ (\sqrt{3})^2 + (1)^2 \right] + 3 \left( \frac{2}{\sqrt{3}} \right)^2 \)
\( = 2 \left( \frac{1}{16} + \frac{1}{16} \right) - (3 + 1) + 3 \times \frac{4}{3} = 2 \times \frac{1}{8} - 4 + 4 = \frac{1}{4} \)

Question. Find the value of x in each of the following :
(i) \( \tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ \)
(ii) \( \cos x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ \)
Answer: (i) \( \tan 3x = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \implies \tan 3x = \frac{1}{2} + \frac{1}{2} = 1 \)
\( \tan 3x = \tan 45^\circ \implies 3x = 45^\circ \implies x = 15^\circ \)
(ii) \( \cos x = \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2} \implies \cos x = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
\( \cos x = \cos 30^\circ \implies x = 30^\circ \)

Question. If \( x = 30^\circ \), verify that
(i) \( \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \)
(ii) \( \sin x = \sqrt{\frac{1 - \cos 2x}{2}} \)
Answer: (i) When \( x = 30^\circ, 2x = 60^\circ \).
\( \text{LHS} = \tan 60^\circ = \sqrt{3} \)
\( \text{RHS} = \frac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{2/3} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \). Verified.
(ii) \( \text{RHS} = \sqrt{\frac{1 - \cos 60^\circ}{2}} = \sqrt{\frac{1 - 1/2}{2}} = \sqrt{\frac{1/2}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
\( \text{LHS} = \sin 30^\circ = \frac{1}{2} \). Verified.

Question. Find the value of \( \theta \) in each of the following :
(i) \( 2\sin 2\theta = \sqrt{3} \)
(ii) \( 2\cos 3\theta = 1 \)
Answer: (i) \( \sin 2\theta = \frac{\sqrt{3}}{2} \implies \sin 2\theta = \sin 60^\circ \implies 2\theta = 60^\circ \implies \theta = 30^\circ \)
(ii) \( \cos 3\theta = \frac{1}{2} \implies \cos 3\theta = \cos 60^\circ \implies 3\theta = 60^\circ \implies \theta = 20^\circ \)

Question. If \( \theta \) is an acute angle and \( \sin \theta = \cos \theta \), find the value of \( 2\tan^2 \theta + \sin^2 \theta - 1 \).
Answer: \( \sin \theta = \cos \theta \implies \frac{\sin \theta}{\cos \theta} = 1 \implies \tan \theta = 1 \implies \theta = 45^\circ \)
Value: \( 2\tan^2 45^\circ + \sin^2 45^\circ - 1 = 2(1)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 - 1 = 2 + \frac{1}{2} - 1 = \frac{3}{2} \)

Question. Using the formula, \( \sin(A - B) = \sin A \cos B - \cos A \sin B \), find the value of \( \sin 15^\circ \).
Answer: Let \( A = 45^\circ, B = 30^\circ \). Then \( A - B = 15^\circ \).
\( \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \); \( 0^\circ < A + B \le 90^\circ ; A > B \), find A and B.
Answer: \( A + B = 60^\circ \) (since \( \tan 60^\circ = \sqrt{3} \))
\( A - B = 30^\circ \) (since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \))
Adding both: \( 2A = 90^\circ \implies A = 45^\circ \)
Subtracting: \( 2B = 30^\circ \implies B = 15^\circ \)

TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES

Complementary angles are pairs of angles whose sum is \( 90^\circ \).

Formulae :

• \( \sin(90^\circ - \theta) = \cos \theta \)
• \( \cos(90^\circ - \theta) = \sin \theta \)
• \( \tan(90^\circ - \theta) = \cot \theta \)
• \( \cot(90^\circ - \theta) = \tan \theta \)
• \( \sec(90^\circ - \theta) = \csc \theta \)
• \( \csc(90^\circ - \theta) = \sec \theta \)

Question. Evaluate \( \frac{\tan 65^\circ}{\cot 25^\circ} \).
Answer: \( \frac{\tan(90^\circ - 25^\circ)}{\cot 25^\circ} = \frac{\cot 25^\circ}{\cot 25^\circ} = 1 \)

Question. Without using trigonometric tables, evaluate the following :
(i) \( \frac{\cos 37^\circ}{\sin 53^\circ} \)
(ii) \( \frac{\sin 41^\circ}{\cos 49^\circ} \)
(iii) \( \frac{\sin 30^\circ 17'}{\cos 59^\circ 43'} \)
Answer: (i) \( \frac{\cos(90^\circ - 53^\circ)}{\sin 53^\circ} = \frac{\sin 53^\circ}{\sin 53^\circ} = 1 \)
(ii) \( \frac{\sin(90^\circ - 49^\circ)}{\cos 49^\circ} = \frac{\cos 49^\circ}{\cos 49^\circ} = 1 \)
(iii) \( \frac{\sin(90^\circ - 59^\circ 43')}{\cos 59^\circ 43'} = \frac{\cos 59^\circ 43'}{\cos 59^\circ 43'} = 1 \)

Question. Without using trigonometric tables evaluate the following :
(i) \( \sin^2 25^\circ + \sin^2 65^\circ \)
(ii) \( \cos^2 13^\circ - \sin^2 77^\circ \)
Answer: (i) \( \sin^2(90^\circ - 65^\circ) + \sin^2 65^\circ = \cos^2 65^\circ + \sin^2 65^\circ = 1 \)
(ii) \( \cos^2(90^\circ - 77^\circ) - \sin^2 77^\circ = \sin^2 77^\circ - \sin^2 77^\circ = 0 \)

Question. Without using trigonometric tables, evaluate the following :
(i) \( \frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 \)
(ii) \( \sec 50^\circ \sin 40^\circ + \cos 40^\circ \csc 50^\circ \)
Answer: (i) \( \frac{\cot(90^\circ - 36^\circ)}{\tan 36^\circ} + \frac{\tan(90^\circ - 70^\circ)}{\cot 70^\circ} - 2 = 1 + 1 - 2 = 0 \)
(ii) \( \sec(90^\circ - 40^\circ) \sin 40^\circ + \cos 40^\circ \csc(90^\circ - 40^\circ) = \csc 40^\circ \sin 40^\circ + \cos 40^\circ \sec 40^\circ = 1 + 1 = 2 \)

Question. Express each of the following in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \);
(i) \( \csc 69^\circ + \cot 69^\circ \)
(ii) \( \sin 81^\circ + \tan 81^\circ \)
(iii) \( \sin 72^\circ + \cot 72^\circ \)
Answer: (i) \( \csc(90^\circ - 21^\circ) + \cot(90^\circ - 21^\circ) = \sec 21^\circ + \tan 21^\circ \)
(ii) \( \sin(90^\circ - 9^\circ) + \tan(90^\circ - 9^\circ) = \cos 9^\circ + \cot 9^\circ \)
(iii) \( \sin(90^\circ - 18^\circ) + \cot(90^\circ - 18^\circ) = \cos 18^\circ + \tan 18^\circ \)

Question. If \( \tan 2\theta = \cot(\theta + 6^\circ) \), find \( \theta \).
Answer: \( \cot(90^\circ - 2\theta) = \cot(\theta + 6^\circ) \implies 90^\circ - 2\theta = \theta + 6^\circ \implies 3\theta = 84^\circ \implies \theta = 28^\circ \)

Question. If A, B, C are the interior angles of a triangle ABC, prove that \( \tan \frac{B+C}{2} = \cot \frac{A}{2} \).
Answer: \( A + B + C = 180^\circ \implies \frac{B+C}{2} = 90^\circ - \frac{A}{2} \)
\( \tan \left(\frac{B+C}{2}\right) = \tan(90^\circ - A/2) = \cot(A/2) \). Proved.

TRIGONOMETRIC IDENTITIES

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• \( 1 + \tan^2 \theta = \sec^2 \theta \)
• \( 1 + \cot^2 \theta = \csc^2 \theta \)

Question. Prove the identity: \( (1 - \sin^2 \theta) \sec^2 \theta = 1 \).
Answer: \( \text{LHS} = \cos^2 \theta \cdot \sec^2 \theta = \cos^2 \theta \cdot \frac{1}{\cos^2 \theta} = 1 = \text{RHS} \)

Question. Prove the identity: \( \frac{\sin \theta}{1 - \cos \theta} = \csc \theta + \cot \theta \).
Answer: Multiply by \( (1 + \cos \theta) \): \( \frac{\sin \theta(1 + \cos \theta)}{1 - \cos^2 \theta} = \frac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta} = \frac{1 + \cos \theta}{\sin \theta} = \csc \theta + \cot \theta \)

Question. If \( \sin \theta + \sin^2 \theta = 1 \), prove that \( \cos^2 \theta + \cos^4 \theta = 1 \).
Answer: \( \sin \theta = 1 - \sin^2 \theta = \cos^2 \theta \).
Now, \( \cos^2 \theta + \cos^4 \theta = \sin \theta + \sin^2 \theta = 1 \). Proved.

Question. Prove the following identities :
(i) \( \cos^4 A - \cos^2 A = \sin^4 A - \sin^2 A \)
(ii) \( \cot^4 A - 1 = \csc^4 A - 2\csc^2 A \)
(iii) \( \sin^6 A + \cos^6 A = 1 - 3\sin^2 A \cos^2 A \).
Answer: (i) We have, \( \text{LHS} = \cos^4 A - \cos^2 A = \cos^2 A (\cos^2 A - 1) \)
\( = -\cos^2 A (1 - \cos^2 A) = -\cos^2 A \sin^2 A \)
\( = -(1 - \sin^2 A) \sin^2 A = -\sin^2 A + \sin^4 A \)
\( = \sin^4 A - \sin^2 A = \text{RHS} \)
(ii) We have, \( \text{LHS} = \cot^4 A - 1 = (\csc^2 A - 1)^2 - 1 \)
[\( \because \cot^2 A = \csc^2 A - 1 \)]
\( \therefore \cot^4 A = (\csc^2 A - 1)^2 \)
\( = \csc^4 A - 2\csc^2 A + 1 - 1 \)
\( = \csc^4 A - 2\csc^2 A = \text{RHS} \)
(iii) We have, \( \text{LHS} = \sin^6 A + \cos^6 A = (\sin^2 A)^3 + (\cos^2 A)^3 \)
\( = (\sin^2 A + \cos^2 A) \{(\sin^2 A)^2 + (\cos^2 A)^2 - \sin^2 A \cos^2 A\} \)
[\( \because a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)]
\( = \{(\sin^2 A)^2 + (\cos^2 A)^2 + 2\sin^2 A \cos^2 A - 3\sin^2 A \cos^2 A\} \)
\( = [(\sin^2 A + \cos^2 A)^2 - 3\sin^2 A \cos^2 A] \)
\( = 1 - 3\sin^2 A \cos^2 A = \text{RHS} \)

Question. Prove the following identities :
(i) \( \frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A \cos^2 A} - 2 \)
(ii) \( \frac{\cos A}{1 - \tan A} + \frac{\sin^2 A}{\sin A - \cos A} = \sin A + \cos A \)
(iii) \( \frac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{\cos^2 \theta} = 2\left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) \)
Answer: (i) We have, \( \text{LHS} = \frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A} = \frac{\sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A} \) [on taking LCM]
\( = \frac{(\sin^2 A)^2 + (\cos^2 A)^2 + 2\sin^2 A \cos^2 A - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \)
\( = \frac{(\sin^2 A + \cos^2 A)^2 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \)
\( = \frac{1 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} = \frac{1}{\sin^2 A \cos^2 A} - 2 = \text{RHS} \)
(ii) We have, \( \text{LHS} = \frac{\cos A}{1 - \tan A} + \frac{\sin^2 A}{\sin A - \cos A} \)
\( = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin^2 A}{\sin A - \cos A} \)
\( = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} \)
\( = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} \)
\( = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A - \sin A} \)
\( = \cos A + \sin A = \text{RHS} \)
(iii) We have, \( \text{LHS} = \frac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{\cos^2 \theta} \)
\( = \frac{(1 + 2\sin \theta + \sin^2 \theta) + (1 - 2\sin \theta + \sin^2 \theta)}{\cos^2 \theta} \)
\( = \frac{2 + 2\sin^2 \theta}{\cos^2 \theta} = \frac{2(1 + \sin^2 \theta)}{1 - \sin^2 \theta} = 2\left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) = \text{RHS} \).

Question. Prove the following identities :
(i) \( 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 = 0 \)
(ii) \( (\sin^8 \theta - \cos^8 \theta) = (\sin^2 \theta - \cos^2 \theta)(1 - 2\sin^2 \theta \cos^2 \theta) \)
Answer: (i) We have, \( \text{LHS} = 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 \)
\( = 2 [(\sin^2 \theta)^3 + (\cos^2 \theta)^3] - [3 ((\sin^2 \theta)^2 + (\cos^2 \theta)^2)] + 1 \)
\( = 2[(\sin^2 \theta + \cos^2 \theta) \{(\sin^2 \theta)^2 + (\cos^2 \theta)^2 - \sin^2 \theta \cos^2 \theta\}] \)
\( - 3[(\sin^2 \theta)^2 + (\cos^2 \theta)^2 + 2\sin^2 \theta \cos^2 \theta - 2\sin^2 \theta \cos^2 \theta] + 1 \)
\( = 2[(\sin^2 \theta)^2 + (\cos^2 \theta)^2 + 2\sin^2 \theta \cos^2 \theta - 3\sin^2 \theta \cos^2 \theta] \)
\( - 3 [(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta] + 1 \)
\( = 2[(\sin^2 \theta + \cos^2 \theta)^2 - 3\sin^2 \theta \cos^2 \theta] - 3[1 - 2\sin^2 \theta \cos^2 \theta] + 1 \)
\( = 2(1 - 3\sin^2 \theta \cos^2 \theta) - 3(1 - 2\sin^2 \theta \cos^2 \theta) + 1 \)
\( = 2 - 6\sin^2 \theta \cos^2 \theta - 3 + 6\sin^2 \theta \cos^2 \theta + 1 \)
\( = 0 = \text{RHS} \)
(ii) We have, \( \text{LHS} = \sin^8 \theta - \cos^8 \theta = (\sin^4 \theta)^2 - (\cos^4 \theta)^2 \)
\( = (\sin^4 \theta - \cos^4 \theta)(\sin^4 \theta + \cos^4 \theta) \)
\( = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta + \cos^4 \theta) \)
\( = (\sin^2 \theta - \cos^2 \theta) \{(\sin^2 \theta)^2 + (\cos^2 \theta)^2 + 2\sin^2 \theta \cos^2 \theta - 2\sin^2 \theta \cos^2 \theta\} \)
\( = (\sin^2 \theta - \cos^2 \theta) \{(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta\} \)
\( = (\sin^2 \theta - \cos^2 \theta) (1 - 2\sin^2 \theta \cos^2 \theta) = \text{RHS} \)

Question. If \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \), prove that each of the side is equal to \( \pm 1 \).
Answer: We have, \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \)
Multiplying both sides by \( (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \) we get
\( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \)
\( = (\sec A - \tan A)^2 (\sec B - \tan B)^2 (\sec C - \tan C)^2 \)
\( \implies (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C) = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2 \)
\( \implies 1 = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2 \)
\( \implies (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) = \pm 1 \)
Similarly, multiplying both sides by \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) \), we get
\( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = \pm 1 \)

Question. If \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), show that \( m^2 - n^2 = 4\sqrt{mn} \).
Answer: We have, \( \text{LHS} = m^2 - n^2 = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 \)
\( = 4 \tan \theta \sin \theta \) [\( \because (a+b)^2 - (a-b)^2 = 4ab \)]
And, \( \text{RHS} = 4\sqrt{mn} \)
\( = 4 \sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)} = 4 \sqrt{\tan^2 \theta - \sin^2 \theta} \)
\( = 4 \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta} = 4 \sqrt{\frac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta}} \)
\( = 4 \sqrt{\frac{\sin^2 \theta(1 - \cos^2 \theta)}{\cos^2 \theta}} = 4 \sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}} = 4 \frac{\sin^2 \theta}{\cos \theta} \)
\( = 4 \sin \theta \frac{\sin \theta}{\cos \theta} = 4 \sin \theta \tan \theta \)
Thus we have \( \text{LHS} = \text{RHS} \), i.e. \( m^2 - n^2 = 4\sqrt{mn} \)

Question. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), show that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).
Answer: We have, \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)
\( \implies (\cos \theta + \sin \theta)^2 = 2 \cos^2 \theta \)
\( \implies \cos^2 \theta + \sin^2 \theta + 2 \cos \theta \sin \theta = 2 \cos^2 \theta \)
\( \implies \cos^2 \theta - 2 \cos \theta \sin \theta = \sin^2 \theta \)
\( \implies \cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta = 2 \sin^2 \theta \)
\( \implies (\cos \theta - \sin \theta)^2 = 2 \sin^2 \theta \)
\( \implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta \)

Question. If \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \csc \theta = q \), show that \( q(p^2 - 1) = 2p \).
Answer: We have, \( \text{LHS} = q(p^2 - 1) = (\sec \theta + \csc \theta) [(\sin \theta + \cos \theta)^2 - 1] \)
\( = \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1\} \)
\( = \left( \frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta} \right) [1 + 2\sin \theta \cos \theta - 1] \)
\( = \left( \frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta} \right) (2 \sin \theta \cos \theta) \)
\( = 2(\sin \theta + \cos \theta) = 2p = \text{RHS} \)

Question. If \( \sec \theta + \tan \theta = p \), show that \( \frac{p^2 - 1}{p^2 + 1} = \sin \theta \).
Answer: We have, \( \text{LHS} = \frac{p^2 - 1}{p^2 + 1} = \frac{(\sec \theta + \tan \theta)^2 - 1}{(\sec \theta + \tan \theta)^2 + 1} \)
\( = \frac{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta - 1}{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta + 1} \)
\( = \frac{(\sec^2 \theta - 1) + \tan^2 \theta + 2\sec \theta \tan \theta}{\sec^2 \theta + 2\sec \theta \tan \theta + (1 + \tan^2 \theta)} \)
\( = \frac{\tan^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta}{\sec^2 \theta + 2\sec \theta \tan \theta + \sec^2 \theta} = \frac{2\tan^2 \theta + 2\sec \theta \tan \theta}{2\sec^2 \theta + 2\sec \theta \tan \theta} \)
\( = \frac{2\tan \theta (\tan \theta + \sec \theta)}{2\sec \theta (\sec \theta + \tan \theta)} = \frac{\tan \theta}{\sec \theta} = \frac{\sin \theta}{\cos \theta \sec \theta} = \sin \theta = \text{RHS} \)

Question. If \( \frac{\cos \alpha}{\cos \beta} = m \) and \( \frac{\cos \alpha}{\sin \beta} = n \), show that \( (m^2 + n^2) \cos^2 \beta = n^2 \).
Answer: \( \text{LHS} = (m^2 + n^2) \cos^2 \beta = \left( \frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta} \right) \cos^2 \beta \)
[\( \because m = \frac{\cos \alpha}{\cos \beta} \) and \( n = \frac{\cos \alpha}{\sin \beta} \)]
\( = \left( \frac{\cos^2 \alpha \sin^2 \beta + \cos^2 \alpha \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \right) \cos^2 \beta \)
\( = \cos^2 \alpha \left( \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \right) \cos^2 \beta = \frac{\cos^2 \alpha}{\sin^2 \beta} = \left( \frac{\cos \alpha}{\sin \beta} \right)^2 = n^2 = \text{RHS} \)

Question. If \( a \cos \theta + b \sin \theta = m \) and \( a \sin \theta - b \cos \theta = n \), prove that \( a^2 + b^2 = m^2 + n^2 \).
Answer: We have, \( \text{RHS} = m^2 + n^2 = (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 \)
\( = (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta) \)
\( = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 = \text{LHS} \).

Question. If \( a \cos \theta - b \sin \theta = c \), prove that \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \).
Answer: We have, \( (a \cos \theta - b \sin \theta)^2 + (a \sin \theta + b \cos \theta)^2 \)
\( = (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) \)
\( = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \)
\( \implies c^2 + (a \sin \theta + b \cos \theta)^2 = a^2 + b^2 \) [\( \because a \cos \theta - b \sin \theta = c \)]
\( \implies (a \sin \theta + b \cos \theta)^2 = a^2 + b^2 - c^2 \)
\( \implies a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \).

Question. Prove that : \( (1 - \sin \theta + \cos \theta)^2 = 2(1 + \cos \theta)(1 - \sin \theta) \)
Answer: \( (1 - \sin \theta + \cos \theta)^2 = 1 + \sin^2 \theta + \cos^2 \theta - 2 \sin \theta + 2 \cos \theta - 2 \sin \theta \cos \theta \)
\( = 2 - 2 \sin \theta + 2 \cos \theta - 2 \sin \theta \cos \theta \)
\( = 2 (1 - \sin \theta) + 2 \cos \theta (1 - \sin \theta) \)
\( = 2(1 - \sin \theta)(1 + \cos \theta) = \text{RHS} \)

Question. If \( \sin \theta + \sin^2 \theta = 1 \), prove that \( \cos^2 \theta + \cos^4 \theta = 1 \).
Answer: We have, \( \sin \theta + \sin^2 \theta = 1 \implies \sin \theta = 1 - \sin^2 \theta \implies \sin \theta = \cos^2 \theta \)
Now, \( \cos^2 \theta + \cos^4 \theta = \cos^2 \theta + (\cos^2 \theta)^2 = \cos^2 \theta + \sin^2 \theta = 1 \)

Question. Prove that : \( \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = \frac{2}{2\sin^2 \theta - 1} \)
Answer: We have, \( \text{LHS} = \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = \frac{(\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)} \)
\( = \frac{2(\sin^2 \theta + \cos^2 \theta)}{\sin^2 \theta - \cos^2 \theta} \) [\( \because (a+b)^2 + (a-b)^2 = 2(a^2 + b^2) \)]
\( = \frac{2}{\sin^2 \theta - (1 - \sin^2 \theta)} = \frac{2}{2\sin^2 \theta - 1} = \text{RHS} \).

Question. Express the ratios \( \cos A, \tan A \) and \( \sec A \) in terms of \( \sin A \).
Answer: Since \( \cos^2 A + \sin^2 A = 1 \), therefore, \( \cos^2 A = 1 - \sin^2 A \), i.e., \( \cos A = \pm \sqrt{1 - \sin^2 A} \). This gives \( \cos A = \sqrt{1 - \sin^2 A} \). Hence, \( \tan A = \frac{\sin A}{\cos A} = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \) and \( \sec A = \frac{1}{\cos A} = \frac{1}{\sqrt{1 - \sin^2 A}} \).

Question. Prove that \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \), using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
Answer: \( \text{LHS} = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} = \frac{(\tan \theta + \sec \theta) - 1}{(\tan \theta - \sec \theta) + 1} \)
\( = \frac{\{(\tan \theta + \sec \theta) - 1\}(\tan \theta - \sec \theta)}{\{(\tan \theta - \sec \theta) + 1\}(\tan \theta - \sec \theta)} = \frac{(\tan^2 \theta - \sec^2 \theta) - (\tan \theta - \sec \theta)}{\{\tan \theta - \sec \theta + 1\}(\tan \theta - \sec \theta)} \)
\( = \frac{-1 - \tan \theta + \sec \theta}{(\tan \theta - \sec \theta + 1)(\tan \theta - \sec \theta)} = \frac{-1}{\tan \theta - \sec \theta} = \frac{1}{\sec \theta - \tan \theta} \), which is the RHS of the identity, we are required to prove.

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