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