Question
Prove the following trigonometric identities.
$\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}=2\text{cosec A}$
$\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}=2\text{cosec A}$
✓
Answer
Consider the L.H.S.
$\text{L.H.S}=\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}$
$=\frac{\tan\text{A}(1-\sec\text{A})-\tan\text{A}(1+\sec\text{A})}{(1+\sec\text{A})(1-\sec\text{A})}$
$=\frac{\tan\text{A}-\tan\text{A}\sec\text{A}-\tan\text{A}-\tan\text{A}\sec\text{A}}{(1-\sec^2\text{A})}$
$=\frac{-2\tan\text{A}\sec\text{A}}{(1-\sec^2\text{A})}$
$=\frac{-2\tan\text{A}\sec\text{A}}{-\tan^2\text{A}}$
$=\frac{2\sec\text{A}}{\tan\text{A}}$
$=2\text{ cosec A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$\text{L.H.S}=\frac{\tan\text{A}}{1+\sec\text{A}}-\frac{\tan\text{A}}{1-\sec\text{A}}$
$=\frac{\tan\text{A}(1-\sec\text{A})-\tan\text{A}(1+\sec\text{A})}{(1+\sec\text{A})(1-\sec\text{A})}$
$=\frac{\tan\text{A}-\tan\text{A}\sec\text{A}-\tan\text{A}-\tan\text{A}\sec\text{A}}{(1-\sec^2\text{A})}$
$=\frac{-2\tan\text{A}\sec\text{A}}{(1-\sec^2\text{A})}$
$=\frac{-2\tan\text{A}\sec\text{A}}{-\tan^2\text{A}}$
$=\frac{2\sec\text{A}}{\tan\text{A}}$
$=2\text{ cosec A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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