General Aptitude TestQuantitative ReasoningTrigonometry

≤ft(sqrtsec ^2 θ+cosec^2 θ)≤ft((sin θ(1+cos θ))/(1+cos θ-sin ^2 θ)),…

$\left(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\right)\left(\frac{\sin \theta(1+\cos \theta)}{1+\cos \theta-\sin ^2 \theta}\right), 0^{\circ}<\theta<90^{\circ}$ is equal to:

Answer & explanation

Correct answer: option 1

( $\sqrt {sec²θ + cosec²θ }$ ) . ( $\frac{ sinθ ( 1 + cosθ)}{1 + cosθ - sin²θ}$ )

{ Using , secθ = $\frac{1}{cosθ}$ , cosecθ = $\frac{1}{sinθ}$ & sin²θ + cos²θ = 1 }

= ( $\sqrt {1/cos²θ + 1/sin²θ }$ ) . ( $\frac{ sinθ ( 1 + cosθ)}{1 + cosθ - 1 + cos²θ}$ )

= ( $\sqrt {(sin²θ + cos²θ)/sin²θ .cos²θ }$ ) . ( $\frac{ sinθ ( 1 + cosθ)}{ cosθ + cos²θ}$ )

= $\frac{1}{sinθ.cosθ}$ . ( $\frac{ sinθ ( 1 + cosθ)}{ cosθ(1 + cosθ)}$ )

= $\frac{1}{sinθ.cosθ}$ . ( $\frac{ sinθ}{ cosθ}$ )

= $\frac{1}{cos²θ}$

= sec²θ