A) 2
B) ? 1
C) \[\frac{a}{b}\]
D) 0
E) \[\frac{b}{a}\]
Correct Answer: B
Solution :
\[y={{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)\] Let \[a=r\sin \theta \]and \[b=r\cos \theta \] \ \[y={{\tan }^{-1}}\left[ \frac{r\sin (\theta -x)}{r\cos (\theta -x)} \right]\] \[y=\theta -x\];\[y={{\tan }^{-1}}\left( \frac{a}{b} \right)-x\] \[\frac{dy}{dx}=-1\].
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