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

If $y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$, then $\frac{d y}{d x}$ is equal to
  1. 0
  2. $\frac{a}{b}$
  3. -1
  4. 2

Solution

$y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$ $ =\tan ^{-1}\left(\frac{\frac{a}{b-\tan x}}{1+\frac{a}{b} \tan x}\right) $ [Take $b \cos x \operatorname{common}$ from numerator and denominator] $ \begin{array}{rlrl} & =\tan ^{-1}\left(\frac{\tan \theta-\tan x}{1+\tan \theta \tan x}\right) & {\left[\text { Let } \frac{a}{b}=\tan \theta\right]} \\ & =\tan ^{-1}(\tan \{\theta-x\})=\theta-x & & {\left[\because \frac{a}{b}=\tan \theta\right]} \\ y & =\tan ^{-1} \frac{a}{b}-x & \\ \Rightarrow \frac{d y}{d x} & =0-1=-1 & \end{array} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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