Question

If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then  } \frac{dy}{dx}\] is equal to ___________ .

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Answer 1

`1`

 

\[\text { We have, y } = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{1 + \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)^2}\frac{d}{dx}\left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\frac{d}{dx}\left( \sin x + \cos x \right) - \left( \sin x + \cos x \right)\frac{d}{dx}\left( \cos x - \sin x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\left( \cos x - \sin x \right) - \left( \sin x + \cos x \right)\left( - \sin x - \cos x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\left( \cos x - \sin x \right) + \left( \sin x + \cos x \right)\left( \sin x + \cos x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2} \times \frac{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}{\left( \cos x - \sin x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = 1\]