Question

$\frac{d}{dx}{\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$-1$

Put $a=r\cos \alpha ,b=r\sin \alpha$

$\therefore r=\sqrt{a^2+b^2},\tan \alpha =\frac{b}{a}$

$\therefore y={\tan }^{-1}\frac{r\cos (x+\alpha )}{r\sin (x+\alpha )}={\tan }^{-1}\cot (x+\alpha )$

$={\tan }^{-1}\tan [\frac{\pi }{2}-(x+\alpha )]$

$\therefore y=\frac{\pi }{2}-x-\alpha$

$\therefore \frac{dy}{dx}=-1$