If \[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] then \[\frac{dy}{dx}=\] [Kerala (Engg.) 2005]

If \[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x}

1.	If \[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] then \[\frac{dy}{dx}=\] [Kerala (Engg.) 2005]
A.	2
B.	? 1
C.	\[\frac{a}{b}\]
D.	0
E.	\[\frac{b}{a}\]
Answer» C. \[\frac{a}{b}\]

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If \[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] then \[\frac{dy}{dx}=\] [Kerala (Engg.) 2005]

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