SSC CGL 2021 PYQ — Question 61

SSC CGL - 17 Aug 2021 - Shift 1 • Quantitative Aptitude • Q61

sec2θ+cosec2θ×Sinθ(1+Cosθ)1+Cosθ−Sin2θ\sqrt{sec^2 \theta + cosec^2 \theta} \times \frac{Sin \theta (1 + Cos \theta)}{1 + Cos \theta - Sin^2 \theta} 0° < θ\theta < 90° is equal to :

Show answer & explanation

Correct answer: A

Explanation

The correct answer is Option 1 i.e sec2θ.

sec2θ+cosec2θ×Sinθ(1+Cosθ)1+Cosθ−Sin2θ\sqrt{sec^2 \theta + cosec^2 \theta} \times \frac{Sin \theta (1 + Cos \theta)}{1 + Cos \theta - Sin^2 \theta} can be written as,

⇒1Cos2θ+1Sin2θ×Sin​​θ(1+Cosθ)(1+Cosθ)(Cosθ\sqrt{\frac{1}{Cos^2 \theta} + \frac{1}{Sin^2 \theta}} \times \frac{ Sin​​ \theta (1 + Cos \theta)}{(1 + Cos \theta)(Cos \theta}

⇒ sec2θ×cosec2θ×SinθCosθ\sqrt{sec^2 \theta \times cosec^2 \theta} \times \frac{Sin \theta }{Cos \theta}

⇒ Secθ\theta × Cosecθ\theta × SinθCosθ \frac{Sin \theta }{Cos \theta}

⇒  Secθ\theta × Cosecθ\theta × SecθCosecθ\frac{Sec \theta}{Cosec \theta}

⇒ Sec2θSec^2 \theta.