Question

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

Answer 1

We have ,

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

$\Rightarrow LHS=\frac{co{t}^{2}A(secA-1)(secA+1)+se{c}^{2}A(sinA-1)(1+sinA)}{(1+sinA)(1+secA)}$

$\Rightarrow LHS=\frac{co{t}^{2}A(se{c}^{2}A-1)+se{c}^{2}A(si{n}^{2}A-1)}{(1+sinA)(1+secA)}$

$\Rightarrow LHS=\frac{co{t}^{2}A(se{c}^{2}A-1)-se{c}^{2}A(1-si{n}^{2}A)}{(1+sinA)(1+secA)}$

$\Rightarrow LHS=\frac{co{t}^{2}Ata{n}^{2}A-se{c}^{2}Aco{s}^{2}A}{(1+sinA)(1+secA)}$

$\Rightarrow LHS=\frac{1-1}{(1+sinA)(1+secA)}=0=RHS$