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प्रश्न
Differentiate \[\tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right), - \pi < x < \pi\] ?
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उत्तर
\[\text{ Let }f\left( x \right) = \tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right)\]
This function is defined for all real numbers where cos x ≠ 1
\[f\left( x \right) = \tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right)\]
\[ \Rightarrow f\left( x \right) = \tan^{- 1} \left[ \frac{2 \sin\left( \frac{x}{2} \right)\cos\left( \frac{x}{2} \right)}{2 \cos^2 \left( \frac{x}{2} \right)} \right]\]
\[ \Rightarrow f\left( x \right) = \tan^{- 1} \left[ \tan\left( \frac{x}{2} \right) \right] = \frac{x}{2}\]
\[\text{ Thus}, f'\left( x \right) = \frac{d}{dx}\left( \frac{x}{2} \right) = \frac{1}{2}\]
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