Question

If $y={\tan }^{-1}\left(\frac{a\sin x+b\cos x}{a\cos x-b\sin x}\right),-\frac{\pi }{2}<x<\frac{\pi }{2}$, $\frac{b}{a}\tan x<1$ then $\frac{dy}{dx}$ equals

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

The correct option is A $1$

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