Question

Prove that (cosec theta-sintheta)(sec theta-cos theta)=(1)÷(tan theta +cot theta)

Answer 1

Step-by-step explanation:

We have,

LHS

$\rm (\cosec θ - \sin θ)(\sec θ-\cos θ)$

$\rm \bigg(\dfrac{1}{\sin θ}- \sin θ \bigg) \bigg(\dfrac{1}{\cos θ}-\cos θ \bigg)$

$\rm \bigg(\dfrac{1- \sin ^2 θ}{\sin θ} \bigg) \bigg(\dfrac{1-\cos ^2 θ}{\cos θ} \bigg)$

$\rm \bigg(\dfrac{\cos ^2 θ}{\sin θ} \bigg) \bigg(\dfrac{\sin ^2 θ}{\cos θ} \bigg)$

$\rm \bigg(\dfrac{\cos ^{\cancel{2}} θ}{\cancel{\sin θ}} \bigg) \bigg(\dfrac{\sin ^{\cancel{2}} θ}{\cancel{\cos θ}} \bigg)$

$\rm \cos θ \sin θ$

RHS

$\rm \dfrac{1}{\tan θ + \cot θ}$

$\rm \dfrac{1}{\frac{\sin θ}{\cos θ} + \frac{\cos \theta}{\sin θ}}$

$\rm \dfrac{\sin θ \cos θ}{ \sin ^2 θ + \cos ^2 θ }$

$\rm \dfrac{\sin θ \cos θ}{ 1 }$

$\rm = \sin θ \cos θ$

$\rm \qquad \qquad LHS=RHS$