Prove the following. 1+ - 1- =2cosecA - Vidyadip

Question

Prove the following.
$\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}=2\text{cosecA}$

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Answer

LHS $=\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}=\frac{\tan\text{A}(1-\sec\text{A}-1-\sec\text{A})}{(1+\sec\text{A})(1-\sec\text{A})}$
$=\frac{\tan\text{A}(-2\sec\text{A})}{(1-\sec^2\text{A})}=\frac{2\tan\text{A}\cdot\sec\text{A}}{(\sec^2\text{A}-1)}$ $[\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$\frac{2\tan\text{A}\cdot\sec\text{A}}{\tan^2\text{A}}[\because\sec^2\text{A}-\tan^2\text{A}=1]$ $\bigg[\because\sec\theta=\frac{1}{\cos\theta}\text{ and }\tan\theta=\frac{\sin\theta}{\cos\theta}\bigg]$
$=\frac{2\sec\text{A}}{\tan\text{A}}=\frac{2}{\sin\text{A}}=2\text{cosecA}=$ RHS $\Big[\because\text{cosec}\theta=\frac{1}{\sin\theta}\Big]$

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