If $y = an^{- 1}{\left( \frac{a\cos{x - b\sin x}}{b\cos{x + a\sin x}} ight),\ then\ \frac{dy}{dx}}$ is equal to

Question

If $y = 	an^{- 1}{\left( \frac{a\cos{x - b\sin x}}{b\cos{x + a\sin x}} ight),\ then\ \frac{dy}{dx}}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\because y = 	an^{- 1}\left( \frac{a\cos{x - b\sin x}}{b\cos{x + a\sin x}} ight)$
$= 	an^{- 1}\left( \frac{\frac{a}{b} - 	an x}{1 + \frac{a}{b}	an x} ight)$
$= 	an^{- 1}\left\lbrack 	an\left( 	an^{- 1}{\left( \frac{a}{b} ight) - x} ight) ightbrack$
$\Rightarrow y = 	an^{- 1}{\left( \frac{a}{b} ight) - x}$
$	herefore\frac{dy}{dx} = 0 - 1 = - 1$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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