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
• E. $\frac{b}{a}$

Answer 1

Answer: (B)

$y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$
$={\tan }^{-1}\left(\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}\right)$
$={\tan }^{-1}\left\{\tan \left({\tan }^{-1}\left(\frac{a}{b}\right)-x\right)\right\}$
$\Rightarrow$ $y={\tan }^{-1}\left(\frac{a}{b}\right)-x$
On differentiating w.r.t., $x,$ we get
$\frac{dy}{dx}=0-1=-1$