A) 2
B) \[-1\]
C) \[\frac{a}{b}\]
D) 0
E) \[\frac{b}{a}\]
Correct Answer: B
Solution :
\[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x} \right)\] \[={{\tan }^{-1}}\left\{ \tan \left( {{\tan }^{-1}}\left( \frac{a}{b} \right)-x \right) \right\}\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left( \frac{a}{b} \right)-x\] On differentiating w.r.t.,\[x,\]we get \[\frac{dy}{dx}=0-1=-1\]
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