2022 Trigonometry Questions and Solutions
1: Given that cos θ = √3/2, then the value of \(\frac{\csc^2 \theta – \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta}\) is [CBSE 2022](1 marks)
(a) -1
(b) 1
(c) 1/2
(d) -1/2
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Solution:
Given: cos θ = √3/2 ⇒ θ = 30°
sin θ = 1/2, csc θ = 2, sec θ = 2/√3
csc²θ = 4, sec²θ = 4/3
Numerator: csc²θ – sec²θ = 4 – 4/3 = 8/3
Denominator: csc²θ + sec²θ = 4 + 4/3 = 16/3
Value = (8/3)/(16/3) = 1/2
Answer: (c) 1/2
2: \(\frac{1}{\csc \theta (1 – \cot \theta)} + \frac{1}{\sec \theta (1 – \tan \theta)}\) is equal to [CBSE 2022](1 marks)
(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ – cos θ
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Solution:
Simplify each term:
First term: \(\frac{1}{\csc \theta (1 – \cot \theta)} = \frac{\sin \theta}{1 – \cot \theta}\)
Second term: \(\frac{1}{\sec \theta (1 – \tan \theta)} = \frac{\cos \theta}{1 – \tan \theta}\)
Convert to sin and cos:
= \(\frac{\sin \theta}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\cos \theta}{1 – \frac{\sin \theta}{\cos \theta}}\)
= \(\frac{\sin \theta}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\cos \theta}{\frac{\cos \theta – \sin \theta}{\cos \theta}}\)
= \(\frac{\sin^2 \theta}{\sin \theta – \cos \theta} + \frac{\cos^2 \theta}{\cos \theta – \sin \theta}\)
= \(\frac{\sin^2 \theta – \cos^2 \theta}{\sin \theta – \cos \theta}\)
= \(\frac{(\sin \theta – \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta – \cos \theta}\)
= sin θ + cos θ
Answer: (c) sin θ + cos θ
3: The value of θ for which 2 sin 2θ = 1, is [CBSE 2022](1 marks)
(a) 15°
(b) 30°
(c) 45°
(d) 60°
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Solution:
2 sin 2θ = 1 ⇒ sin 2θ = 1/2
We know sin 30° = 1/2, so 2θ = 30° ⇒ θ = 15°
Also sin 150° = 1/2, but 2θ = 150° ⇒ θ = 75° (not in options)
Answer: (a) 15°
4: If sin²θ + sinθ = 1, then find the value of cos²θ + cos⁴θ is [CBSE 2022]( 1 marks)
(a) -1
(b) 1
(c) 0
(d) 2
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Solution:
Given: sin²θ + sinθ = 1 ⇒ sin²θ = 1 – sinθ
We know sin²θ + cos²θ = 1 ⇒ cos²θ = 1 – sin²θ = sinθ
Now cos²θ + cos⁴θ = cos²θ + (cos²θ)²
= sinθ + (sinθ)² = sinθ + sin²θ
= 1 (from given equation)
Answer: (b) 1