Question
Prove that
$\frac{\cot A-1}{2-\sec ^2 A}=\frac{\cot A}{1+\tan A}$
$\frac{\cot A-1}{2-\sec ^2 A}=\frac{\cot A}{1+\tan A}$
✓
Answer
$\text { L.H.S. }=\frac{\cot A -1}{2-\sec ^2 A } $
$ =\frac{\frac{1}{\tan A }-1}{2-\left(1+\tan ^2 A \right)}$
$=\frac{1-\tan A }{\tan A \left(1-\tan ^2 A \right)} $
$ =\frac{1-\tan A }{\tan A (1-\tan A )(1+\tan A )}$
$ =\frac{1}{\tan A (1+\tan A )}$
$=\frac{1}{\tan A } \times \frac{1}{1+\tan A } $
$ =\frac{\cot A }{1+\tan A }$
$R.H.S.$
Hence proved.
$ =\frac{\frac{1}{\tan A }-1}{2-\left(1+\tan ^2 A \right)}$
$=\frac{1-\tan A }{\tan A \left(1-\tan ^2 A \right)} $
$ =\frac{1-\tan A }{\tan A (1-\tan A )(1+\tan A )}$
$ =\frac{1}{\tan A (1+\tan A )}$
$=\frac{1}{\tan A } \times \frac{1}{1+\tan A } $
$ =\frac{\cot A }{1+\tan A }$
$R.H.S.$
Hence proved.
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