(1+cosθ-sin²θ)/(sinθ(1+cosθ)) × (√(sec²θ+cosec²θ))/(tanθ+cotθ) , 0º < θ < 90º, is equal to:

(1+cosθ-sin²θ)/(sinθ(1+cosθ)) ×…

$\frac{1+cosθ-sin^2θ}{sinθ(1+cosθ)}$ × $\frac{\sqrt {sec^2θ+cosec^2θ}}{tanθ+cotθ}$ , 0º < θ < 90º, is equal to:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. cot θ

= $\frac{1+cosθ-sin^2θ}{sinθ(1+cosθ)}$ × $\frac{\sqrt {sec^2θ+cosec^2θ}}{tanθ+cotθ}$

= (1 - sin² $\theta$ + cos $\theta$ )/(sin $\theta$ (1 + cos $\theta$ ) × (1/sin $\theta$ cos $\theta$ )/(1/sin $\theta$ cos $\theta$ )

= cos $\theta$ /sin $\theta$ × 1

= cot $\theta$ .