SarkariPYQ

SSC CHSL 2021 PYQ — Question 67

SSC CHSL - 19 April 2021 - Shift 3 • Quantitative Aptitude • Q67

Back to paper

If sec⁡(3θ2)=2,0°<θ<90°,\sec({{3\theta} \over 2})=\sqrt2,0°<\theta<90°,then the value  (2sinθ+4cos2θ)\ ({{2sin\theta+4cos2\theta}}) will be:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is Option 1 i.e. 3.

sec⁡(3θ2)=2\sec({{3\theta} \over 2})=\sqrt2

 sec⁡(3θ2)=2\sec({{3\theta} \over 2})=\sqrt2 = sec⁡45°\sec 45°

3θ = 45º × 2 = 90º

θ = 30º

Now,  (2sinθ+4cos2θ)\ ({{2sin\theta+4cos2\theta}}) = 2 × sin 30º + 4 × cos 60º

= 2 × (1/2) + 4 × (1/2) = 1 + 2 = 3