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