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If $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$, then $\frac{d y}{d x}$
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The correct answer is:
-1
$$
y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}
$$
Put $a=r \cos \alpha, b=r \sin \alpha$
$$
\begin{aligned}
& \therefore y=\tan ^{-1}\left\{\frac{r(\cos x \cos \alpha-\sin x \sin \alpha)}{r(\sin \alpha \cos x+\cos \alpha \sin x)}\right\} \\
& =\tan ^{-1}\left[\frac{\cos (x+\alpha)}{\sin (x+\alpha)}\right]=\tan ^{-1}[\cot (x+\alpha)] \\
& =\tan ^{-1}\left\{\tan \left[\frac{\pi}{2}-(x+\alpha)\right]\right\}=\frac{\pi}{2}-(x+\alpha) \\
& \therefore \frac{d y}{d x}=0-(1+0)=-1
\end{aligned}
$$
y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}
$$
Put $a=r \cos \alpha, b=r \sin \alpha$
$$
\begin{aligned}
& \therefore y=\tan ^{-1}\left\{\frac{r(\cos x \cos \alpha-\sin x \sin \alpha)}{r(\sin \alpha \cos x+\cos \alpha \sin x)}\right\} \\
& =\tan ^{-1}\left[\frac{\cos (x+\alpha)}{\sin (x+\alpha)}\right]=\tan ^{-1}[\cot (x+\alpha)] \\
& =\tan ^{-1}\left\{\tan \left[\frac{\pi}{2}-(x+\alpha)\right]\right\}=\frac{\pi}{2}-(x+\alpha) \\
& \therefore \frac{d y}{d x}=0-(1+0)=-1
\end{aligned}
$$
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