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Question
Prove the following identitie
Solution
\[\text{ LHS }= \frac{\tan x}{1 - \cot x} + \frac{\cot x}{1 - \tan x}\]
\[ = \frac{\frac{\sin x}{\cos x}}{1 - \frac{\cos x}{\sin x}} + \frac{\frac{\cos x}{\sin x}}{1 - \frac{\sin x}{\cos x}}\]
\[ = \frac{\frac{\sin x}{\cos x}}{\frac{\sin x - \cos x}{\sin x}} + \frac{\frac{\cos x}{\sin x}}{\frac{\cos x - \sin x}{\cos x}}\]
\[ = \frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x - \cos x} + \frac{\cos x}{\sin x} \times \frac{\cos x}{\cos x - \sin x}\]
\[ = \frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x - \cos x} + \frac{\cos x}{\sin x} \times \frac{\cos x}{- \left( \sin x - \cos x \right)}\]
\[ = \frac{\sin^2 x}{\cos x\left( \sin x - \cos x \right)} - \frac{\cos^2 x}{\sin x\left( \sin x - \cos x \right)}\]
\[ = \frac{\sin^3 x - \cos^3 x}{\sin x \cos x\left( \sin x - \cos x \right)}\]
\[ = \frac{\left( \sin x - \cos x \right)\left[ \sin^2 x + \cos^2 x + \sin x\cos x \right]}{\sin x \cos x\left( \sin x - \cos x \right)}\]
\[ = \frac{1 \times \left[ 1 + \sin x \cos x \right]}{\sin x \cos x}\]
\[ = \frac{1 + \sin x \cos x}{\sin x \cos x}\]
\[ = \frac{1}{\sin x \cos x} + \frac{\sin x \cos x}{\sin x \cos x}\]
\[ = \frac{1}{\sin x} \times \frac{1}{\cos x} + 1\]
\[ = cosec x \times \sec x + 1\]
\[ = \left( \sec x co\sec x + 1 \right)\]
= RHS
Hence proved.
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