Question

If $ y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right), $ then $ \frac{dy}{dx} $ is equal to:

• A. 2
• B. $ -1 $
• C. $ \frac{a}{b} $
• D. 0
• E. $ \frac{b}{a} $

Answer 1

Answer: (B)

$ y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right) $
$ ={{\tan }^{-1}}\left( \frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x} \right) $
$ ={{\tan }^{-1}}\left\{ \tan \left( {{\tan }^{-1}}\left( \frac{a}{b} \right)-x \right) \right\} $
$ \Rightarrow $ $ y={{\tan }^{-1}}\left( \frac{a}{b} \right)-x $
On differentiating w.r.t., $ x, $ we get
$ \frac{dy}{dx}=0-1=-1 $