question_answer

Differentiate \[{{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right),\] \[\frac{-\,\pi }{2}<x<\frac{\pi }{2}\] and \[\frac{a}{b}\tan x>-\,1\,\,w.r.t.x.\]

Answer:

Let \[y={{\tan }^{-\,1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] \[\Rightarrow \]   \[y={{\tan }^{-\,1}}\left( \frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x} \right)\] \[\Rightarrow \]   \[y={{\tan }^{-\,1}}\left( \frac{a}{b} \right)-{{\tan }^{-\,1}}(\tan x)\] \[\Rightarrow \]   \[y={{\tan }^{-\,1}}\left( \frac{a}{b} \right)-x\]            \[\left[ \because -\frac{\pi }{2}<x<\frac{\pi }{2} \right]\] \[\Rightarrow \]   \[\frac{dy}{dx}=0-1=-\,1\]