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Prove the following trigonometric identity :
$\frac{\tan \theta}{1 + \cot \theta} + \frac{\cot \theta}{1 + \tan \theta} = \tan \theta + \cot \theta - 1$
$\frac{\tan \theta}{1 + \cot \theta} + \frac{\cot \theta}{1 + \tan \theta} = \tan \theta + \cot \theta - 1$
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Solution: $LHS = \frac{\tan \theta}{1 + \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 + \tan \theta}$
$= \frac{\tan^2 \theta}{1 + \tan \theta} + \frac{1}{\tan \theta(1 + \tan \theta)}$
$= \frac{1 + \tan^3 \theta}{\tan \theta(1 + \tan \theta)}$
$= \frac{(1 + \tan \theta)(1 + \tan^2 \theta - \tan \theta)}{\tan \theta(1 + \tan \theta)}$
$= \cot \theta + \tan \theta - 1 = RHS$
$= \frac{\tan^2 \theta}{1 + \tan \theta} + \frac{1}{\tan \theta(1 + \tan \theta)}$
$= \frac{1 + \tan^3 \theta}{\tan \theta(1 + \tan \theta)}$
$= \frac{(1 + \tan \theta)(1 + \tan^2 \theta - \tan \theta)}{\tan \theta(1 + \tan \theta)}$
$= \cot \theta + \tan \theta - 1 = RHS$