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Prove that : $\frac{\cot A - 1}{2 - \sec^2 A} = \frac{\cot A}{1 + \tan A}$
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$LHS = \frac{\frac{1}{\tan A} - 1}{2 - (1+\tan^2 A)}$ ($1$ mark)
$= \frac{\frac{1 - \tan A}{\tan A}}{2 - 1 - \tan^2 A}$ ($1/2$ mark)
$= \frac{1 - \tan A}{\tan A(1 - \tan^2 A)}$ ($1/2$ mark)
$= \frac{1}{\tan A (1 + \tan A)}$ ($1/2$ mark)
$= \frac{\cot A}{1 + \tan A} = RHS$ ($1/2$ mark)
$= \frac{\frac{1 - \tan A}{\tan A}}{2 - 1 - \tan^2 A}$ ($1/2$ mark)
$= \frac{1 - \tan A}{\tan A(1 - \tan^2 A)}$ ($1/2$ mark)
$= \frac{1}{\tan A (1 + \tan A)}$ ($1/2$ mark)
$= \frac{\cot A}{1 + \tan A} = RHS$ ($1/2$ mark)