Question:
$\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \operatorname{cosec} A$
Solution:
$\mathrm{LHS}=\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=\frac{\tan A(1-\sec A-1-\sec A)}{(1+\sec A)(1-\sec A)}$
$=\frac{\tan A(-2 \sec A)}{\left(1-\sec ^{2} A\right)}=\frac{2 \tan A \cdot \sec A}{\left(\sec ^{2} A-1\right)}$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\frac{2 \tan A \cdot \sec A}{\tan ^{2} A}$ $\left[\because \sec ^{2} A-\tan ^{2} A=1\right]\left[\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{2 \sec A}{\tan A}=\frac{2}{\sin A}=2 \operatorname{cosec} A=$ RHS $\left[\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$