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)

$-1$

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