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
    $0$
  • B
    $\frac{a}{b}$
  • C
    $-1$
  • D
    $2$

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Similar Questions

The differential coefficient of ${\tan ^{ - 1}}\left( {\frac{x}{{1 + \sqrt {1 - {x^2}} }}} \right)$ with respect to ${\sin ^{ - 1}}x$ is:

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $

If $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$,then $\frac{d y}{d x}=$

$\frac{d}{dx} \tan^{-1} \left( \frac{1-x}{1+x} \right) = $ . . . . . .

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ has the value

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