Question

If $y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right),\frac{a}{b}\tan x>-1,$ then absolute value of $\frac{dy}{dx}$ is equal to

Answer 1

$y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)={\tan }^{-1}\left(\frac{\frac{a}{b}-\frac{\sin x}{\cos x}}{1+\frac{a}{b}\cdot \frac{\sin x}{\cos x}}\right)={\tan }^{-1}\left(\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}\right)(\because \frac{a}{b}\tan x>-1)y={\tan }^{-1}\frac{a}{b}-{\tan }^{-1}\left(\tan x\right)y={\tan }^{-1}\frac{a}{b}-(x+k\pi )(\text{where}k\in Z)\Rightarrow \frac{dy}{dx}=-1$
$\Rightarrow \left|\frac{dy}{dx}\right|=1$