2024 Trigonometry Questions and Solutions
1. If \(\sin \theta = \frac{1}{3}\), then \(\sec \theta\) is equal to: ( CBSE 2024)
(a) \(\frac{2\sqrt{2}}{3}\)
(b) \(\frac{3}{2\sqrt{2}}\)
(c) 3
(d) \(\frac{1}{\sqrt{3}}\)
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Solution: Given: \(\sin \theta = \frac{1}{3}\)
Using identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 – \sin^2 \theta = 1 – \left(\frac{1}{3}\right)^2 = 1 – \frac{1}{9} = \frac{8}{9}\)
\(\cos \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\)
\(\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{2\sqrt{2}}{3}} = \frac{3}{2\sqrt{2}}\)
Answer: (b) \(\frac{3}{2\sqrt{2}}\)
2. For what value of \(\theta\), \(\sin^2 \theta + \sin \theta + \cos^2 \theta\) is equal to 2? (CBSE 2024)
(a) \(45^\circ\)
(b) \(0^\circ\)
(c) \(90^\circ\)
(d) \(30^\circ\)
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Solution: \(\sin^2 \theta + \sin \theta + \cos^2 \theta\)
\(= (\sin^2 \theta + \cos^2 \theta) + \sin \theta = 1 + \sin \theta\)
We need: \(1 + \sin \theta = 2\) So
\(\sin \theta = 1\)
This occurs when \(\theta = 90^\circ\)
Answer: (c) \(90^\circ\)
3. In a \(\Delta ABC\), \(\angle A = 90^\circ\). If \(\tan C = \sqrt{3}\), then find the value of \(\sin B + \cos C – \cos^2 B\). (CBSE 2024)
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Solution:Given: \(\angle A = 90^\circ\) and \(\tan C = \sqrt{3}\)
We know \(\tan 60^\circ = \sqrt{3}\),
so \(\angle C = 60^\circ\)
Since \(\angle A + \angle B + \angle C = 180^\circ\)
\(90^\circ + \angle B + 60^\circ = 180^\circ\)
\(\angle B = 30^\circ\) Now,
\(\sin B + \cos C – \cos^2 B\)
\(= \sin 30^\circ + \cos 60^\circ – \cos^2 30^\circ\)
\(= \frac{1}{2} + \frac{1}{2} – \left(\frac{\sqrt{3}}{2}\right)^2 = 1 – \frac{3}{4} = \frac{1}{4}\)
Answer: \(\frac{1}{4}\)
4. Prove that: \(\sqrt{\frac{\sec A – 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A – 1}} = 2 \csc A\) (CBSE 2024)
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Solution: Let \(LHS = \sqrt{\frac{\sec A – 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A – 1}}\)
Rationalizing both terms:
\(= \frac{\sqrt{\sec A – 1}}{\sqrt{\sec A + 1}} \times \frac{\sqrt{\sec A – 1}}{\sqrt{\sec A – 1}} \)+ \(\frac{\sqrt{\sec A + 1}}{\sqrt{\sec A – 1}} \times \frac{\sqrt{\sec A + 1}}{\sqrt{\sec A + 1}}\)
\(= \frac{\sec A – 1}{\sqrt{\sec^2 A – 1}} + \frac{\sec A + 1}{\sqrt{\sec^2 A – 1}}\)
\(= \frac{(\sec A – 1) + (\sec A + 1)}{\sqrt{\sec^2 A – 1}} = \frac{2\sec A}{\sqrt{\sec^2 A – 1}}\)
We know \(\sec^2 A – 1 = \tan^2 A\),
so: \(= \frac{2\sec A}{\tan A} = 2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A}\) \(= 2 \times \frac{1}{\sin A} = 2 \csc A\) Hence proved.
5. If \(\sec \theta + \tan \theta = m\), then the value of \(\sec \theta – \tan \theta\) is: (CBSE 2024)
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6. If \(\cos (\alpha + \beta) = 0\), then value of \(\cos \left( \frac{\alpha + \beta}{2} \right)\) is equal to: (CBSE 2024)
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{1}{2}\)
(c) 0
(d) \(\sqrt{2}\)
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Solution: Given: \(\cos (\alpha + \beta) = 0\) We know \(\cos 90^\circ = 0\), so \(\alpha + \beta = 90^\circ\) Then \(\frac{\alpha + \beta}{2} = \frac{90^\circ}{2} = 45^\circ\) Now, \(\cos \left( \frac{\alpha + \beta}{2} \right) = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
Answer: (a) \(\frac{1}{\sqrt{2}}\)
7. (A) Evaluate: \(2\sqrt{2} \cos 45^\circ \sin 30^\circ + 2\sqrt{3} \cos 30^\circ\) (CBSE 2024)
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Solution: We know: \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), \(\sin 30^\circ = \frac{1}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
Substituting these values: \(2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2\sqrt{3} \times \frac{\sqrt{3}}{2}\) \(= 2 \times \frac{1}{2} + 2 \times \frac{3}{2}\) \(= 1 + 3 = 4\)
Answer: 4
OR
Verify: \(\sin (A + B)\) \(= \sin A \cos B + \cos A \sin B\) for \(A = 60^\circ\), \(B = 30^\circ\) (CBSE 2024)
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Solution:Left side: \(\sin (60^\circ + 30^\circ) = \sin 90^\circ = 1\)
Right side: \(\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\)
\(= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}\)
\(= \frac{3}{4} + \frac{1}{4} = 1\)
∴ LHS = RHS, hence verified.
8. Prove that: \(\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta\) (CBSE 2024)
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Solution:Left side: \(\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta}\)
Express in terms of sin and cos: \(= \frac{\frac{\sin \theta}{\cos \theta}}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 – \frac{\sin \theta}{\cos \theta}}\)
\(= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta – \sin \theta}{\cos \theta}}\)
\(= \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta – \cos \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta – \sin \theta}\)
\(= \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta (\sin \theta – \cos \theta)}\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{\sin^2 \theta}{\cos \theta} – \frac{\cos^2 \theta}{\sin \theta} \right)\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{\sin^3 \theta – \cos^3 \theta}{\sin \theta \cos \theta} \right)\) Using identity: \(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta} \right)\)
\(= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta}{\sin \theta \cos \theta}\)
\(= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}\) (since \(\sin^2 \theta + \cos^2 \theta = 1\))
\(= \frac{1}{\sin \theta \cos \theta} + 1 = \csc \theta \sec \theta + 1\)
\(= 1 + \sec \theta \csc \theta\) Hence proved.