Question

If $y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$, then $\frac{dy}{dx}$ is equal to

• A. $2$
• B. $-1$
• C. $\frac{a}{b}$
• D. $0$

Answer 1

Answer: (B)

$-1$

$y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$
Let $a=r\sin \theta ;b=r\cos \theta$
$\therefore y={\tan }^{-1}\left\{\frac{r\sin (\theta -x)}{r\cos (\theta -x)}\right\}$
$\Rightarrow y=\theta -x,y={\tan }^{-1}\left(\frac{a}{b}\right)-x$
$\therefore \frac{dy}{dx}=-1$