Question

Question 2
$\frac{tanA}{1+secA}-\frac{tanA}{1-secA}=2cosecA$

Answer 1

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

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

$=\frac{2tanA.secA}{ta{n}^{2}A}[\because se{c}^{2}A-ta{n}^{2}A=1]$

$[\because sec\theta =\frac{1}{cos\theta }andtan\theta =\frac{sin\theta }{cos\theta }]$

$=\frac{2secA}{tanA}=\frac{2}{sinA}=2cosecA=RHS[\because cosec\theta =\frac{1}{sin\theta }]$