Question

Prove the following questions :
$\frac{tanA}{1+secA}-\frac{tanA}{1-secA}=2cosecA$

Answer 1

We will start from LHS .

[We can take $tanA$ as common and $(1-secA0(1+secA)$ makes $(1-se{c}^{2}A=ta{n}^{2}A)$ so formulae will be used.]

LHS $=\frac{tanA}{1+secA}-\frac{tanA}{1-secA}$

$=tanA[\frac{1}{(1+secA)}-\frac{1}{(1-secA)}]$ [Taking $tanA$ common]

$=tanA\left[\frac{1-secA-(1+secA)}{(1+secA)(1-secA)}\right]$ {Taking LCM)

$=tanA\left[\frac{-2secA}{(1-se{c}^{2}A)}\right]=\frac{tanA(-2secA)}{-(se{c}^{2}A-1)}$

$=\frac{-tanA2secA}{-ta{n}^{2}A}=\frac{2secA}{tanA}$

$=\frac{2\times \frac{1}{cosA}}{\frac{sinA}{cosA}}$

$=\frac{2}{sinA}=2cosecA=$ RHS

Hence , proved .