Question

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

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

Answer 1

The correct option is B $1$

$y={\tan }^{-1}\left(\frac{a\sin x-b\cos x}{a\cos x+b\sin x}\right)={\tan }^{-1}\left(\frac{a\cos x(\tan x-\frac{b}{a})}{a\cos x(1+\frac{b}{a}\tan x)}\right)$
Let $\frac{b}{a}=\tan \theta$
$\therefore y={\tan }^{-1}\left(\frac{\tan x-\tan \theta }{1+\tan \theta \tan x}\right)y={\tan }^{-1}\tan (x-\theta )=x-\theta$
$\frac{dy}{dx}=1-0=1$