Question
Prove that $\frac{\cot A}{1-\cot A}+\frac{\tan A}{1-\tan A}=-1$
✓
Answer
$\text { L.H.S }=\frac{\cot A}{1-\cot A}+\frac{\tan A}{1-\tan A}$
$=\frac{\cot A}{1-\frac{1}{\tan A}}+\frac{\tan A}{1-\tan A}$
$=\frac{\cot A}{\frac{\tan A-1}{\tan A}}+\frac{\tan A}{1-\tan A}$
$=\frac{\cot A \tan A}{\tan A-1}+\frac{\tan A}{1-\tan A}$
$=\frac{1}{\tan A-1}+\frac{\tan A}{1-\tan A} \quad \ldots \ldots[\because \cot A \tan A=1]$
$=-\frac{1}{1-\tan A}+\frac{\tan A}{1-\tan A}$
$=-\left(\frac{1}{1-\tan A}-\frac{\tan A}{1-\tan A}\right)$
$=-\left(\frac{1-\tan A }{1-\tan A }\right)$
$=-1$
$=\text { R.H.S }$
$\therefore \frac{\cot A}{1-\cot A}+\frac{\tan A}{1-\tan A}=-1$
$=\frac{\cot A}{1-\frac{1}{\tan A}}+\frac{\tan A}{1-\tan A}$
$=\frac{\cot A}{\frac{\tan A-1}{\tan A}}+\frac{\tan A}{1-\tan A}$
$=\frac{\cot A \tan A}{\tan A-1}+\frac{\tan A}{1-\tan A}$
$=\frac{1}{\tan A-1}+\frac{\tan A}{1-\tan A} \quad \ldots \ldots[\because \cot A \tan A=1]$
$=-\frac{1}{1-\tan A}+\frac{\tan A}{1-\tan A}$
$=-\left(\frac{1}{1-\tan A}-\frac{\tan A}{1-\tan A}\right)$
$=-\left(\frac{1-\tan A }{1-\tan A }\right)$
$=-1$
$=\text { R.H.S }$
$\therefore \frac{\cot A}{1-\cot A}+\frac{\tan A}{1-\tan A}=-1$
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