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प्रश्न
Differentiate
\[\tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right), \frac{\pi}{4} < x < \frac{\pi}{4}\] ?
उत्तर
\[\text{ Let, y} = \tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{\cos x + \sin x}{\cos x}}{\frac{\cos x - \sin x}{\cos x}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{1 + \tan x}{1 - \tan x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} + x \right) \right]\]
\[ \Rightarrow y = \frac{\pi}{4} + x\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = 0 + 1\]
\[ \therefore \frac{d y}{d x} = 1\]
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