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Question
Prove the following identities
\[\frac{\cos x}{1 - \sin x} = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}\]
Solution
\[\text{ RHS }= \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}\]
\[ = \frac{\left( 1 + \cos x \right) + \left( \sin x \right)}{\left( 1 + cosx \right) - \left( \sin x \right)}\]
\[ = \frac{\left[ \left( 1 + \cos x \right) + \left( \sin x \right) \right]\left[ \left( 1 + \cos x \right) + \left( \sin x \right) \right]}{\left[ \left( 1 + \cos x \right) - \left( \sin x \right) \right]\left[ \left( 1 + \cos x \right) + \left( \sin x \right) \right]}\]
\[ = \frac{\left[ \left( 1 + \cos x \right) + \left( \sin x \right) \right]^2}{\left( 1 + \cos x \right)^2 - \left( \sin x \right)^2}\]
\[ = \frac{\left( 1 + \cos x \right)^2 + \left( \sin x \right)^2 + 2\left( 1 + \cos x \right)\left( \sin x \right)}{1^2 + \cos^2 x + 2 \times 1 \times \cos x - \sin^2 x}\]
\[ = \frac{1 + \cos^2 x + 2\cos x + \sin^2 x + 2\sin x \cos x + 2\sin x}{1 + \cos^2 x + 2\cos x - \sin^2 x}\]
\[ = \frac{1 + \left( \sin^2 x + \cos^2 x \right) + 2\cos x + 2\sin x \cos x + 2\sin x}{\left( 1 - \sin^2 x \right) + \cos^2 x + 2\cos x}\]
\[ = \frac{1 + 1 + 2\cos x + 2\sin x \cos x + 2\sin x}{\cos^2 x + \cos^2 x + 2\cos x}\]
\[ = \frac{2 + 2\cos x + 2\sin x \cos x + 2\sin x}{2 \cos^2 x + 2\cos x}\]
\[ = \frac{1 + \cos x + \sin x \cos x + \sin x}{\cos^2 x + \cos x}\]
\[ = \frac{1\left( 1 + \cos x \right) + \sin x\left( \cos x + 1 \right)}{\cos x\left( \cos x + 1 \right)}\]
\[ = \frac{\left( \cos x + 1 \right)\left( 1 + \sin x \right)}{\cos x\left( cosx + 1 \right)}\]
\[ = \frac{\left( 1 + \sin x \right)}{\cos x}\]
\[ = \frac{\left( 1 + \sin x \right) \times \cos x}{\cos x \times \cos x}\]
\[ = \frac{\left( 1 + \sin x \right) \times \cos x}{\cos^2 x}\]
\[ = \frac{\left( 1 + \sin x \right) \times \cos x}{1 - \sin^2 x}\]
\[ = \frac{\left( 1 + \sin x \right) \times \cos x}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}\]
\[ = \frac{\cos x}{1 - \sin x}\]
= LHS
Hence proved.
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