CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set 03

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Very Short Answer 

Question. If \( \tan \theta + \cot \theta = 5 \), find the value of \( \tan^2 \theta + \cot^2 \theta \).
Answer: Sol. \( \tan \theta + \cot \theta = 5 \) ...[Given
\( \tan^2 \theta + \cot^2 \theta + 2 \tan \theta \cot \theta = 25 \)
...[Squaring both sides
\( \tan^2 \theta + \cot^2 \theta + 2 = 25 \)
\( \therefore \tan^2 \theta + \cot^2 \theta = 23 \)

Question. If \( \sec 2A = \text{cosec}(A - 27^\circ) \) where 2A is an acute angle, find the measure of A.
Answer: Sol. \( \sec 2A = \text{cosec}(A - 27^\circ) \)
\( \text{cosec}(90^\circ - 2A) = \text{cosec}(A - 27^\circ) \)
...[\( \sec \theta = \text{cosec}(90^\circ - \theta) \)]
\( 90^\circ - 2A = A - 27^\circ \)
\( 90^\circ + 27^\circ = 2A + A \)
\( \Rightarrow 3A = 117^\circ \)
\( \therefore \angle A = \frac{117^\circ}{3} = 39^\circ \)

Question. If \( \tan \alpha = \sqrt{3} \) and \( \tan \beta = \frac{1}{\sqrt{3}} \), \( 0 < \alpha, \beta < 90^\circ \), find the value of \( \cot(\alpha + \beta) \).
Answer: Sol. \( \tan \alpha = \sqrt{3} = \tan 60^\circ \) ...(i)
\( \tan \beta = \frac{1}{\sqrt{3}} = \tan 30^\circ \) ...(ii)
Solving (i) & (ii), \( \alpha = 60^\circ \) and \( \beta = 30^\circ \)
\( \therefore \cot(\alpha + \beta) = \cot(60^\circ + 30^\circ) = \cot 90^\circ = 0 \)

Question. If \( \sin \theta - \cos \theta = 0 \), find the value of \( \sin^4 \theta + \cos^4 \theta \).
Answer: Sol. \( \sin \theta - \cos \theta = 0 \Rightarrow \sin \theta = \cos \theta \)
\( \Rightarrow \frac{\sin \theta}{\cos \theta} = 1 \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^\circ \)
Now, \( \sin^4 \theta + \cos^4 \theta \)
\( = \sin^4 45^\circ + \cos^4 45^\circ \)
\( = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)

Question. If \( \sec \theta + \tan \theta = 7 \), then evaluate \( \sec \theta - \tan \theta \).
Answer: Sol. We know that,
\( \sec^2 \theta - \tan^2 \theta = 1 \)
\( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \)
\( (7)(\sec \theta - \tan \theta) = 1 \)
...[\( \sec \theta + \tan \theta = 7 \); (Given)]
\( \therefore \sec \theta - \tan \theta = \frac{1}{7} \)

Question. Evaluate: \( 10 \cdot \frac{1 - \cot^2 45^\circ}{1 + \sin^2 90^\circ} \).
Answer: Sol. \( 10 \cdot \frac{1 - \cot^2 45^\circ}{1 + \sin^2 90^\circ} = 10 \cdot \frac{1 - (1)^2}{1 + (1)^2} = 10 \cdot \left(\frac{0}{2}\right) = 0 \)

Question. If \( \text{cosec } \theta = \frac{5}{4} \), find the value of \( \cot \theta \).
Answer: Sol. We know that, \( \cot^2 \theta = \text{cosec}^2 \theta - 1 \)
\( = \left(\frac{5}{4}\right)^2 - 1 \Rightarrow \frac{25}{16} - 1 \Rightarrow \frac{25 - 16}{16} \)
\( \cot^2 \theta = \frac{9}{16} \therefore \cot \theta = \frac{3}{4} \)

Question. If \( \theta = 45^\circ \), then what is the value of \( 2 \sec^2 \theta + 3 \text{cosec}^2 \theta \)?
Answer: Sol. \( 2 \sec^2 \theta + 3 \text{cosec}^2 \theta = 2 \sec^2 45^\circ + 3 \text{cosec}^2 45^\circ \)
\( = 2(\sqrt{2})^2 + 3(\sqrt{2})^2 = 4 + 6 = 10 \)

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

Question. Evaluate: \( \sin^2 19^\circ + \sin^2 71^\circ \).
Answer: Sol. \( \sin^2 19^\circ + \sin^2 71^\circ \)
\( = \sin^2 19^\circ + \sin^2(90^\circ - 19^\circ) \) ...[\( \sin(90^\circ - \theta) = \cos \theta \)]
\( = \sin^2 19^\circ + \cos^2 19^\circ = 1 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]

Question. What happens to value of \( \cos \theta \) when \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \)?
Answer: Sol. \( \cos 0^\circ = 1, \cos 90^\circ = 0 \)
When \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the value of \( \cos \theta \) decreases from 1 to 0.

Short Answer-I 

Question. Evaluate: \( \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 60^\circ \cdot \tan 65^\circ \cdot \tan 75^\circ - \tan 30^\circ \).
Answer: Sol. \( \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 60^\circ \cdot \tan 65^\circ \cdot \tan 75^\circ - \tan 30^\circ \)
\( = \tan(90^\circ - 75^\circ) \tan(90^\circ - 65^\circ) \cdot \sqrt{3} \cdot \tan 65^\circ \cdot \tan 75^\circ - \frac{1}{\sqrt{3}} \)
\( = \cot 75^\circ \cdot \cot 65^\circ \cdot \sqrt{3} \cdot \frac{1}{\cot 65^\circ} \cdot \frac{1}{\cot 75^\circ} - \frac{1}{\sqrt{3}} \)
...[\( \tan(90^\circ - A) = \cot A \)]
\( = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \)

Question. Express \( \cot 75^\circ + \text{cosec } 75^\circ \) in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 30^\circ \).
Answer: Sol. \( \cot 75^\circ + \text{cosec } 75^\circ \)
\( = \cot(90^\circ - 15^\circ) + \text{cosec}(90^\circ - 15^\circ) \)
\( = \tan 15^\circ + \sec 15^\circ \)
...[\( \cot(90^\circ - A) = \tan A, \text{cosec}(90^\circ - A) = \sec A \)]

Question. If \( \cos(A + B) = 0 \) and \( \sin(A - B) = \frac{1}{2} \), then find the value of A and B where A and B are acute angles.
Answer: Sol. \( \cos(A + B) = 0 \Rightarrow \cos(A + B) = \cos 90^\circ \Rightarrow A + B = 90^\circ \)
\( \sin(A - B) = \frac{1}{2} \Rightarrow \sin(A - B) = \sin 30^\circ \Rightarrow A - B = 30^\circ \)
\( \therefore A = 30^\circ + B \) ...(i)
Substituting (i) in first equation: \( 30^\circ + B + B = 90^\circ \Rightarrow 2B = 60^\circ \Rightarrow B = 30^\circ \)
Putting value of B in (i), we get \( A = 30^\circ + 30^\circ = 60^\circ \)
\( \therefore A = 60^\circ, B = 30^\circ \)

Question. If A, B and C are the interior angles of a \( \Delta ABC \), show that \( \sin\left(\frac{A + B}{2}\right) = \cos\left(\frac{C}{2}\right) \).
Answer: Sol. In \( \Delta ABC \), \( \angle A + \angle B + \angle C = 180^\circ \) ...[Angle sum property of \( \Delta \)]
\( \angle A + \angle B = 180^\circ - \angle C \)
\( \frac{\angle A + \angle B}{2} = \frac{180^\circ - \angle C}{2} = 90^\circ - \frac{\angle C}{2} \) ...(i)
L.H.S. \( = \sin\left(\frac{\angle A + \angle B}{2}\right) = \sin\left(90^\circ - \frac{\angle C}{2}\right) \) ...[From (i)]
\( = \cos\frac{\angle C}{2} = \text{R.H.S.} \)
\( \therefore \text{L.H.S.} = \text{R.H.S.} \)

Question. If \( x = p \sec \theta + q \tan \theta \) and \( y = p \tan \theta + q \sec \theta \), then prove that \( x^2 - y^2 = p^2 - q^2 \).
Answer: Sol. L.H.S. \( = x^2 - y^2 \)
\( = (p \sec \theta + q \tan \theta)^2 - (p \tan \theta + q \sec \theta)^2 \)
\( = p^2 \sec^2 \theta + q^2 \tan^2 \theta + 2pq \sec \theta \tan \theta - (p^2 \tan^2 \theta + q^2 \sec^2 \theta + 2pq \sec \theta \tan \theta) \)
\( = p^2 \sec^2 \theta + q^2 \tan^2 \theta + 2pq \sec \theta \tan \theta - p^2 \tan^2 \theta - q^2 \sec^2 \theta - 2pq \sec \theta \tan \theta \)
\( = p^2(\sec^2 \theta - \tan^2 \theta) - q^2(\sec^2 \theta - \tan^2 \theta) \)
\( = p^2 - q^2 \) ...[\( \sec^2 \theta - \tan^2 \theta = 1 \)]
\( = \text{R.H.S.} \)

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

Question. Simplify: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} \).
Answer: Sol. 1st method:
\( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \)
2nd method:
\( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{\sin^2 A}{\cos^2 A}}{1 + \frac{\cos^2 A}{\sin^2 A}} = \frac{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}}{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} \)
...[\( \sin^2 A + \cos^2 A = 1 \)]
\( = \tan^2 A \)

Question. If \( x = a \cos \theta - b \sin \theta \) and \( y = a \sin \theta + b \cos \theta \), then prove that \( a^2 + b^2 = x^2 + y^2 \).
Answer: Sol. R.H.S. \( = x^2 + y^2 \)
\( = (a \cos \theta - b \sin \theta)^2 + (a \sin \theta + b \cos \theta)^2 \)
\( = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta \)
\( = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) \)
\( = a^2 + b^2 = \text{L.H.S.} \) ...[\( \cos^2 \theta + \sin^2 \theta = 1 \)]

Short Answer-II 

Question. Given \( 2 \cos 3\theta = \sqrt{3} \), find the value of \( \theta \).
Answer: Sol. \( 2 \cos 3\theta = \sqrt{3} \) ...[Given
\( \cos 3\theta = \frac{\sqrt{3}}{2} \Rightarrow \cos 3\theta = \cos 30^\circ \)
\( 3\theta = 30^\circ \therefore \theta = 10^\circ \)

Question. If \( \cos x = \cos 40^\circ \cdot \sin 50^\circ + \sin 40^\circ \cdot \cos 50^\circ \), then find the value of x.
Answer: Sol. \( \cos x = \cos 40^\circ \sin 50^\circ + \sin 40^\circ \cos 50^\circ \)
\( \cos x = \cos 40^\circ \sin(90^\circ - 40^\circ) + \sin 40^\circ \cos(90^\circ - 40^\circ) \)
\( \cos x = \cos^2 40^\circ + \sin^2 40^\circ \)
\( \cos x = 1 \) ...[\( \cos^2 A + \sin^2 A = 1 \)]
\( \cos x = \cos 0^\circ \therefore x = 0^\circ \)

Question. If \( \sin \theta = \frac{1}{2} \), then show that \( 3 \cos \theta - 4 \cos^3 \theta = 0 \).
Answer: Sol. \( \sin \theta = \frac{1}{2} \Rightarrow \sin \theta = \sin 30^\circ \Rightarrow \theta = 30^\circ \)
L.H.S. \( = 3 \cos \theta - 4 \cos^3 \theta \)
\( = 3 \cos 30^\circ - 4 \cos^3(30^\circ) \)
\( = 3\left(\frac{\sqrt{3}}{2}\right) - 4\left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{2} - 4\left(\frac{3\sqrt{3}}{8}\right) \)
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 = \text{R.H.S.} \)

Question. If \( 5 \sin \theta = 4 \), prove that \( \frac{1}{\cos \theta} + \frac{1}{\cot \theta} = 3 \).
Answer: Sol. Given: \( 5 \sin \theta = 4 \Rightarrow \sin \theta = 4/5 \)
\( P/H = 4/5 \). Here, \( P = 4K, H = 5K \).
In rt. \( \Delta ABC \), \( P^2 + B^2 = H^2 \Rightarrow (4K)^2 + B^2 = (5K)^2 \Rightarrow B^2 = 9K^2 \Rightarrow B = 3K \).
\( \sec \theta = \frac{1}{\cos \theta} = \frac{H}{B} = \frac{5K}{3K} = \frac{5}{3} \).
\( \tan \theta = \frac{1}{\cot \theta} = \frac{P}{B} = \frac{4K}{3K} = \frac{4}{3} \).
L.H.S. \( = \frac{1}{\cos \theta} + \frac{1}{\cot \theta} = \sec \theta + \tan \theta = \frac{5}{3} + \frac{4}{3} = \frac{9}{3} = 3 = \text{R.H.S.} \)

Question. Evaluate: \( \frac{\sec 41^\circ \cdot \sin 49^\circ + \cos 29^\circ \cdot \text{cosec } 61^\circ - \frac{2}{\sqrt{3}}(\tan 20^\circ \cdot \tan 60^\circ \cdot \tan 70^\circ)}{3(\sin^2 31^\circ + \sin^2 59^\circ)} \).
Answer: Sol. \( \frac{\sec 41^\circ \sin 49^\circ + \cos 29^\circ \text{cosec } 61^\circ - \frac{2}{\sqrt{3}}(\tan 20^\circ \tan 60^\circ \tan 70^\circ)}{3(\sin^2 31^\circ + \sin^2 59^\circ)} \)
\( = \frac{\sec 41^\circ \sin(90^\circ - 41^\circ) + \cos(90^\circ - 61^\circ) \text{cosec } 61^\circ - \frac{2}{\sqrt{3}}(\tan 20^\circ \cdot \sqrt{3} \cdot \tan(90^\circ - 20^\circ))}{3(\sin^2 31^\circ + \sin^2(90^\circ - 31^\circ))} \)
\( = \frac{\left(\frac{1}{\cos 41^\circ} \cdot \cos 41^\circ\right) + \left(\sin 61^\circ \cdot \frac{1}{\sin 61^\circ}\right) - \frac{2}{\sqrt{3}}(\tan 20^\circ \cdot \sqrt{3} \cdot \cot 20^\circ)}{3(\sin^2 31^\circ + \cos^2 31^\circ)} \)
...[\( \sec \theta = \frac{1}{\cos \theta}, \sin(90^\circ - \theta) = \cos \theta, \text{cosec } \theta = \frac{1}{\sin \theta}, \cos(90^\circ - \theta) = \sin \theta, \tan(90^\circ - \theta) = \cot \theta, \sin^2 \theta + \cos^2 \theta = 1 \)]
\( = \frac{1 + 1 - 2(\tan 20^\circ \cot 20^\circ)}{3(1)} = \frac{2 - 2(1)}{3} = \frac{0}{3} = 0 \)

Question. Evaluate: \( \frac{\sin(50^\circ + \theta) - \cos(40^\circ - \theta)}{\sin 40^\circ \cdot \text{cosec } 40^\circ} + \frac{\tan 1^\circ \cdot \tan 40^\circ \cdot \tan 50^\circ \cdot \tan 89^\circ}{4(\cos^2 29^\circ + \cos^2 61^\circ)} \).
Answer: Sol. \( \frac{\sin(50^\circ + \theta) - \cos(40^\circ - \theta)}{\sin 40^\circ \text{cosec } 40^\circ} + \frac{\tan 1^\circ \tan 40^\circ \tan 50^\circ \tan 89^\circ}{4(\cos^2 29^\circ + \cos^2 61^\circ)} \)
\( = \frac{\sin[90^\circ - (40^\circ - \theta)] - \cos(40^\circ - \theta)}{\sin 40^\circ \cdot \frac{1}{\sin 40^\circ}} + \frac{\tan 1^\circ \tan 40^\circ \tan(90^\circ - 40^\circ) \tan(90^\circ - 1^\circ)}{4(\cos^2 29^\circ + \cos(90^\circ - 29^\circ)^2)} \)
...[\( 50^\circ + \theta = 90^\circ - 40^\circ + \theta = 90^\circ - (40^\circ - \theta) \)]
\( = \frac{\cos(40^\circ - \theta) - \cos(40^\circ - \theta)}{1} + \frac{\tan 1^\circ \tan 40^\circ \cot 40^\circ \cot 1^\circ}{4[\cos^2 29^\circ + \sin^2 29^\circ]} \)
...[\( \sin(90^\circ - A) = \cos A, \tan(90^\circ - A) = \cot A, \tan A = \frac{1}{\cot A}, \sin^2 A + \cos^2 A = 1 \)]
\( = 0 + \frac{1}{4(1)} = \frac{1}{4} \)

Question. Find the value of: \( \left[\frac{\tan 20^\circ}{\text{cosec } 70^\circ}\right]^2 + \left[\frac{\cot 20^\circ}{\sec 70^\circ}\right]^2 + 2\tan 75^\circ \tan 45^\circ \tan 15^\circ \).
Answer: Sol. \( \left[\frac{\tan 20^\circ}{\text{cosec } 70^\circ}\right]^2 + \left[\frac{\cot 20^\circ}{\sec 70^\circ}\right]^2 + 2\tan 75^\circ \tan 45^\circ \tan 15^\circ \)
\( = \left[\frac{\tan(90^\circ - 70^\circ)}{\text{cosec } 70^\circ}\right]^2 + \left[\frac{\cot(90^\circ - 70^\circ)}{\sec 70^\circ}\right]^2 + 2 \tan(90^\circ - 15^\circ) \cdot 1 \cdot \tan 15^\circ \)
\( = \left[\frac{\cot 70^\circ}{\text{cosec } 70^\circ}\right]^2 + \left[\frac{\tan 70^\circ}{\sec 70^\circ}\right]^2 + 2 \cot 15^\circ \cdot \frac{1}{\cot 15^\circ} \)
...[\( \tan(90^\circ - A) = \cot A, \cot(90^\circ - A) = \tan A, \tan A = \frac{1}{\cot A} \)]
\( = \left[\frac{\frac{\cos 70^\circ}{\sin 70^\circ}}{\frac{1}{\sin 70^\circ}}\right]^2 + \left[\frac{\frac{\sin 70^\circ}{\cos 70^\circ}}{\frac{1}{\cos 70^\circ}}\right]^2 + 2 \)
\( = \cos^2 70^\circ + \sin^2 70^\circ + 2 = 1 + 2 = 3 \) ...[\( \cos^2 A + \sin^2 A = 1 \)]

Question. Prove that: \( \frac{\sin^2 63^\circ + \sin^2 27^\circ}{\sec^2 20^\circ - \cot^2 70^\circ} + 2 \sin 36^\circ \sin 42^\circ \sec 48^\circ \sec 54^\circ = 3 \).
Answer: Sol. \( \frac{\sin^2 63^\circ + \sin^2 27^\circ}{\sec^2 20^\circ - \cot^2 70^\circ} + 2 \sin 36^\circ \sin 42^\circ \sec 48^\circ \sec 54^\circ \)
\( = \frac{\sin^2 63^\circ + \sin^2(90^\circ - 63^\circ)}{\sec^2 20^\circ - \cot^2(90^\circ - 20^\circ)} + 2 \sin(90^\circ - 54^\circ) \sin(90^\circ - 48^\circ) \sec 48^\circ \sec 54^\circ \)
\( = \frac{\sin^2 63^\circ + \cos^2 63^\circ}{\sec^2 20^\circ - \tan^2 20^\circ} + 2 \cos 54^\circ \cos 48^\circ \cdot \frac{1}{\cos 48^\circ} \cdot \frac{1}{\cos 54^\circ} \)
...[\( \sin(90^\circ - A) = \cos A, \cot(90^\circ - A) = \tan A, \sin^2 A + \cos^2 A = 1, \sec^2 A - \tan^2 A = 1, \sec A = \frac{1}{\cos A} \)]
\( = \frac{1}{1} + 2 = 3 \)

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: Sol. L.H.S. \( = \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} \)
\( = \frac{(\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta + \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta}{\sin^2 \theta - \cos^2 \theta} \)
\( = \frac{1 + 1}{\sin^2 \theta - (1 - \sin^2 \theta)} = \frac{2}{\sin^2 \theta - 1 + \sin^2 \theta} \)
\( = \frac{2}{2 \sin^2 \theta - 1} = \text{R.H.S.} \) ...(Hence proved)

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

Question. If \( \tan \theta = \frac{a}{b} \), prove that \( \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = \frac{a^2 - b^2}{a^2 + b^2} \).
Answer: Sol. L.H.S. \( = \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} \)
...[Dividing num. and deno. by \( \cos \theta \)]
\( = \frac{a \frac{\sin \theta}{\cos \theta} - b \frac{\cos \theta}{\cos \theta}}{a \frac{\sin \theta}{\cos \theta} + b \frac{\cos \theta}{\cos \theta}} = \frac{a \tan \theta - b}{a \tan \theta + b} \)
\( = \frac{a(\frac{a}{b}) - b}{a(\frac{a}{b}) + b} = \frac{\frac{a^2 - b^2}{b}}{\frac{a^2 + b^2}{b}} = \frac{a^2 - b^2}{a^2 + b^2} = \text{R.H.S.} \)

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

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

Question. Prove that: \( \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta \).
Answer: Sol. L.H.S. \( = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)} \)
\( = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta [2(1 - \sin^2 \theta) - 1]} \) ...[\( \cos^2 \theta = 1 - \sin^2 \theta \)]
\( = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 - 2 \sin^2 \theta - 1)} = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (1 - 2 \sin^2 \theta)} = \tan \theta = \text{R.H.S.} \) ...(Hence proved)

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

Long Answer 

Question. In an acute angled triangle ABC, if \( \sin(A + B - C) = \frac{1}{2} \) and \( \cos(B + C - A) = \frac{1}{2} \), find A, B and C.
Answer: Sol. \( \sin(A + B - C) = \frac{1}{2} = \sin 30^\circ \Rightarrow A + B - C = 30^\circ \) ...(i)
\( \cos(B + C - A) = \frac{1}{2} = \cos 45^\circ \Rightarrow B + C - A = 45^\circ \) ...(ii)
\( A + B + C = 180^\circ \) ...(iii) ...[Sum of all angles of a \( \Delta = 180^\circ \)]
Solving (iii) & (i): \( 2C = 150^\circ \Rightarrow C = 75^\circ \).
Solving (iii) & (ii): \( 2A = 135^\circ \Rightarrow A = 67.5^\circ \).
Putting values of A and C in (iii): \( 67.5^\circ + B + 75^\circ = 180^\circ \Rightarrow B = 180^\circ - 142.5^\circ = 37.5^\circ \).
\( \therefore A = 67.5^\circ, B = 37.5^\circ \text{ and } C = 75^\circ \)

Question. Evaluate: \( \frac{4}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} - \cos^2 45^\circ \).
Answer: Sol. \( \frac{4}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} - \cos^2 45^\circ = 4 \tan^2 30^\circ + \text{cosec}^2 60^\circ - \cos^2 45^\circ \)
\( = 4\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{2}{\sqrt{3}}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{4}{3} + \frac{4}{3} - \frac{1}{2} = \frac{8 + 8 - 3}{6} = \frac{13}{6} \)

Question. Evaluate the following: \( \frac{2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 60^\circ)}{\tan 30^\circ + \cot 60^\circ} \).
Answer: Sol. \( \frac{2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 60^\circ)}{\tan 30^\circ + \cot 60^\circ} \)
\( = \frac{2\left[\left(\frac{1}{\sqrt{2}}\right)^2 + (\sqrt{3})^2\right] - 6\left[\left(\frac{1}{\sqrt{2}}\right)^2 - (\sqrt{3})^2\right]}{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}} = \frac{2\left(\frac{1}{2} + 3\right) - 6\left(\frac{1}{2} - 3\right)}{\frac{2}{\sqrt{3}}} = \frac{2\left(\frac{7}{2}\right) - 6\left(-\frac{5}{2}\right)}{\frac{2}{\sqrt{3}}} = \frac{7 + 15}{\frac{2}{\sqrt{3}}} = 22 \times \frac{\sqrt{3}}{2} = 11\sqrt{3} \)

Question. If \( \theta = 30^\circ \), verify the following:
(i) \( \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \)
(ii) \( \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \).
Answer: Sol. (i) \( \theta = 30^\circ \). L.H.S. \( = \cos(3 \times 30^\circ) = \cos 90^\circ = 0 \). R.H.S. \( = 4 \cos^3 30^\circ - 3 \cos 30^\circ = 4\left(\frac{\sqrt{3}}{2}\right)^3 - 3\left(\frac{\sqrt{3}}{2}\right) = 4\frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \). L.H.S. = R.H.S.
(ii) L.H.S. \( = \sin(3 \times 30^\circ) = \sin 90^\circ = 1 \). R.H.S. \( = 3 \sin 30^\circ - 4 \sin^3 30^\circ = 3(1/2) - 4(1/2)^3 = 3/2 - 4/8 = 3/2 - 1/2 = 1 \). L.H.S. = R.H.S.

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \), where \( 0 < A + B < 90^\circ, A > B \), find A and B. Also calculate: \( \tan A \cdot \sin(A + B) + \cos A \cdot \tan(A - B) \).
Answer: Sol. \( \tan(A + B) = \tan 60^\circ \Rightarrow A + B = 60^\circ \) ...(i); \( \tan(A - B) = \tan 30^\circ \Rightarrow A - B = 30^\circ \) ...(ii). Adding: \( 2A = 90^\circ \Rightarrow A = 45^\circ \). From (i), \( B = 15^\circ \).
Calculation: \( \tan 45^\circ \cdot \sin 60^\circ + \cos 45^\circ \cdot \tan 30^\circ = 1 \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{6}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{12}} = \frac{3\sqrt{3} + \sqrt{6}}{6} \)

Question. If \( \tan(20^\circ - 3\alpha) = \cot(5\alpha - 20^\circ) \), then find the value of \( \alpha \) and hence evaluate: \( \sin \alpha \cdot \sec \alpha \cdot \tan \alpha - \text{cosec } \alpha \cdot \cos \alpha \cdot \cot \alpha \).
Answer: Sol. \( \tan(20^\circ - 3\alpha) = \tan[90^\circ - (5\alpha - 20^\circ)] \Rightarrow 20^\circ - 3\alpha = 110^\circ - 5\alpha \Rightarrow 2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ \).
Evaluation: \( \sin 45^\circ \sec 45^\circ \tan 45^\circ - \text{cosec } 45^\circ \cos 45^\circ \cot 45^\circ = (1/\sqrt{2})(\sqrt{2})(1) - (\sqrt{2})(1/\sqrt{2})(1) = 1 - 1 = 0 \).

Question. If \( \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \) and \( \frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta = 1 \), prove that \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \).
Answer: Sol. Squaring and adding the two equations: \( \left(\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta\right)^2 + \left(\frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta\right)^2 = 1^2 + 1^2 \)
\( \frac{x^2}{a^2} \cos^2 \theta + \frac{y^2}{b^2} \sin^2 \theta + 2\frac{xy}{ab} \cos \theta \sin \theta + \frac{x^2}{a^2} \sin^2 \theta + \frac{y^2}{b^2} \cos^2 \theta - 2\frac{xy}{ab} \sin \theta \cos \theta = 2 \)
\( \frac{x^2}{a^2}(\cos^2 \theta + \sin^2 \theta) + \frac{y^2}{b^2}(\sin^2 \theta + \cos^2 \theta) = 2 \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]

Question. If \( \sqrt{3} \cot^2 \theta - 4 \cot \theta + \sqrt{3} = 0 \), then find the value of \( \cot^2 \theta + \tan^2 \theta \).
Answer: Sol. \( \sqrt{3} \cot^2 \theta - 4 \cot \theta + \sqrt{3} = 0 \Rightarrow \cot \theta + \tan \theta = \frac{4}{\sqrt{3}} \).
Squaring both sides, \( \cot^2 \theta + \tan^2 \theta + 2 \cot \theta \tan \theta = \frac{16}{3} \)
\( \cot^2 \theta + \tan^2 \theta + 2 = \frac{16}{3} \Rightarrow \cot^2 \theta + \tan^2 \theta = \frac{16}{3} - 2 = \frac{10}{3} \)

Question. Prove that \( b^2x^2 - a^2y^2 = a^2b^2 \), if:
(i) \( x = a \sec \theta, y = b \tan \theta \)
(ii) \( x = a \text{cosec } \theta, y = b \cot \theta \).
Answer: Sol. (i) L.H.S. \( = b^2(a \sec \theta)^2 - a^2(b \tan \theta)^2 = b^2a^2 \sec^2 \theta - a^2b^2 \tan^2 \theta = a^2b^2(\sec^2 \theta - \tan^2 \theta) = a^2b^2(1) = a^2b^2 = \text{R.H.S.} \)
(ii) L.H.S. \( = b^2(a \text{cosec } \theta)^2 - a^2(b \cot \theta)^2 = b^2a^2 \text{cosec}^2 \theta - a^2b^2 \cot^2 \theta = a^2b^2(\text{cosec}^2 \theta - \cot^2 \theta) = a^2b^2(1) = a^2b^2 = \text{R.H.S.} \)

Question. If \( \sec \theta - \tan \theta = x \), show that \( \sec \theta + \tan \theta = \frac{1}{x} \) and hence find the values of \( \cos \theta \) and \( \sin \theta \).
Answer: Sol. Part I: \( \sec^2 \theta - \tan^2 \theta = 1 \Rightarrow (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \Rightarrow (\sec \theta + \tan \theta)(x) = 1 \Rightarrow \sec \theta + \tan \theta = 1/x \).
Part II: Adding the two equations: \( 2 \sec \theta = x + 1/x = \frac{x^2+1}{x} \Rightarrow \sec \theta = \frac{x^2+1}{2x} \Rightarrow \cos \theta = \frac{2x}{x^2+1} \).
\( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left[\frac{2x}{x^2+1}\right]^2 = \frac{(x^2+1)^2 - 4x^2}{(x^2+1)^2} = \frac{(x^2-1)^2}{(x^2+1)^2} \Rightarrow \sin \theta = \frac{x^2-1}{x^2+1} \).

Question. If \( \text{cosec } \theta + \cot \theta = p \), then prove that \( \cos \theta = \frac{p^2 - 1}{p^2 + 1} \).
Answer: Sol. \( \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = p \Rightarrow \frac{1 + \cos \theta}{\sin \theta} = p \).
Squaring both sides: \( \frac{(1 + \cos \theta)^2}{\sin^2 \theta} = p^2 \Rightarrow \frac{(1 + \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = p^2 \Rightarrow \frac{1 + \cos \theta}{1 - \cos \theta} = p^2 \)
\( 1 + \cos \theta = p^2 - p^2 \cos \theta \Rightarrow \cos \theta(1 + p^2) = p^2 - 1 \Rightarrow \cos \theta = \frac{p^2 - 1}{p^2 + 1} \) ...Hence proved

Question. If \( \tan \theta + \sin \theta = p \) and \( \tan \theta - \sin \theta = q \), prove that \( p^2 - q^2 = 4\sqrt{pq} \).
Answer: Sol. L.H.S. \( = p^2 - q^2 = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 = 4 \tan \theta \sin \theta \).
R.H.S. \( = 4\sqrt{pq} = 4\sqrt{\tan^2 \theta - \sin^2 \theta} = 4\sqrt{\sin^2 \theta(1/\cos^2 \theta - 1)} = 4\sqrt{\sin^2 \theta \tan^2 \theta} = 4 \tan \theta \sin \theta \).
From (i) and (ii), L.H.S. = R.H.S.

Question. If \( \sin \theta + \cos \theta = m \) and \( \sec \theta + \text{cosec } \theta = n \), then prove that \( n(m^2 - 1) = 2m \).
Answer: Sol. \( m^2 - 1 = (\sin \theta + \cos \theta)^2 - 1 = 1 + 2 \sin \theta \cos \theta - 1 = 2 \sin \theta \cos \theta \).
L.H.S. \( = n(m^2 - 1) = (\sec \theta + \text{cosec } \theta)(2 \sin \theta \cos \theta) = \left(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right) 2 \sin \theta \cos \theta \)
\( = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \cdot 2 \sin \theta \cos \theta = 2(\sin \theta + \cos \theta) = 2m = \text{R.H.S.} \)

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

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