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Question
Differentiate \[\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}\] ?
Solution
\[\text{Let } y = \log\sqrt{\frac{1 - \cos x}{1 + \cos x}}\]
\[ \Rightarrow y = \log \left( \frac{1 - \cos x}{1 + \cos x} \right)^\frac{1}{2} \]
\[ \Rightarrow y = \frac{1}{2}\log\left( \frac{1 - \cos x}{1 + \cos x} \right) \left[ \text{using} \log a^b = b\log a \right]\]
\[\text{Differentiate it with respect to x we get}, \]
\[\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{1}{2}\log\left( \frac{1 - \cos x}{1 + \cos x} \right) \right\}\]
\[ = \frac{1}{2} \times \frac{1}{\left( \frac{1 - \cos x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{1 - \cos x}{1 + \cos x} \right) \left[ \text{Using chain rule} \right]\]
\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\frac{d}{dx}\left( 1 - \cos x \right) - \left( 1 - \cos x \right)\frac{d}{dx}\left( 1 + \cos x \right)}{\left( 1 + cos x \right)^2} \right] \left[ \text{Using quotient rule} \right]\]
\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\left( \sin x \right) - \left( 1 - \cos x \right)\left( - \sin x \right)}{\left( 1 + \cos x \right)^2} \right]\]
\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{\left( 1 + \cos x \right)^2} \right]\]
\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{2\sin x}{\left( 1 + \cos x \right)^2} \right]\]
\[ = \frac{\sin x}{\left( 1 - \cos x \right)\left( 1 + \cos x \right)}\]
\[ = \frac{\sin x}{1 - \cos^2 x}\]
\[ = \frac{\sin x}{\sin^2 x} \]
\[ = \frac{1}{\sin x}\]
`= "cosec "x`
\[So, \frac{d}{dx}\left( \log\sqrt{\frac{1 - \cos x}{1 + \cos x}} \right) = cosec x\]
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