If $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$, then $\frac{d y}{d x}$

If $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$, then $\frac{d y}{d x}$
  1. $\frac{1}{1+x^2}$
  2. $\frac{1}{\sqrt{1-x^2}}$
  3. -1
  4. None of these

Solution

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

Asked in: MHT CET 2021 (21 Sep Shift 2)

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