Question

Prove that :
$(tanA+cosecB{)}^2-(cotB-secA{)}^2=2tanAcotB(cosecA+secB)$

Answer 1

We have,
LHS = $(tanA+cosecB$)${}^2$ $-(cotB-secA$)${}^2$

$\Rightarrow$ LHS = ($ta{n}^{2}A$ + $cose{c}^{2}B$ $+2tanA\cdot cosecB)-$($co{t}^2$ B + $se{c}^2$ A - $2cotB\cdot secA)$

$\Rightarrow$ LHS = ($ta{n}^{2}A$ - $se{c}^{2}A$ )+($cose{c}^2$ B - $co{t}^2$ B)$+2tanA\cdot cosecB+2cotB\cdot secA$

$\Rightarrow$ LHS $=-1+1+2tanA\cdot cosecB+2cotB\cdot secA$

$\Rightarrow$ LHS $=2(tanA\cdot cosecB+cotB\cdot secA)$

$\Rightarrow$ LHS = 2 tan A cot B $(\frac{cosecB}{cotB}+\frac{secA}{tanA})$ [Dividing and multiplying by tan A cot B]

$\Rightarrow$ LHS = 2 tan A cot B $\{\frac{\frac{1}{sinB}}{\frac{cosB}{sinB}}+\frac{\frac{1}{cosA}}{\frac{sinA}{ocsA}}\}$

$\Rightarrow$ LHS = 2 tan A cot B $(\frac{1}{cosB}+\frac{1}{sinA})$ $=2tanAcotB(secB+cosecA)=$ RHS