Question

Prove that $\frac{co{t}^{2}A(secA-1)}{1+sinA}=se{c}^{2}A\left(\frac{1-sinA}{1+secA}\right)$

Answer 1

LHS $=\frac{{\cot }^{2}A(secA-1)}{1+sinA}$

$=\frac{{\cos }^{2}A(1-cosA)}{{\sin }^{2}A(1+sinA)\times cosA}$

$=\frac{cosA(1-cosA)}{{\sin }^{2}A(1+sinA)}\times \frac{(1-sinA)\times (1+cosA)}{(1-sinA)(1+cosA)}$

$=\frac{(1-sinA)}{(1+cosA)}\times \frac{1}{cosA}$

$=\frac{1\times (1-sinA)}{cosA\times cosA(1+secA)}$

$={\sec }^{2}A\left(\frac{1-sinA}{1+secA}\right)$

$=RHS$

Hence proved