If $y = \sec^{- 1}{\lbrack cosec\ x\rbrack +}{cosec}^{- 1}\left\lbrack \sec x \right\rbrack + \sin^{- 1}{\lbrack\cos{x\rbrack + \cos^{- 1}{\lbrack\sin{x\rbrack,\ \ }}}}\ then\frac{dy}{dx}\ is\ equal$ To

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $y = \sec^{- 1}{(cosec\ x) + {cosec}^{- 1}(\sec{x)}} + \sin^{- 1}{(\cos{x) + \cos^{- 1}{(\sin{x)}}}}$
$= \sec^{- 1}{\left\lbrack \sec\left( \frac{\pi}{2} - x \right) \right\rbrack + cosec^{- 1}\left\lbrack cosec\left( \frac{\pi}{2} - x \right) \right\rbrack} + \sin^{- 1}{\left\lbrack \sin\left( \frac{\pi}{2} - x \right) \right\rbrack + {\ \ cos}^{- 1}\left\lbrack \cos\left( \frac{\pi}{2} - x \right) \right\rbrack}$
$= \frac{\pi}{2} - x + \frac{\pi}{2} - x + \frac{\pi}{2} - x + \frac{\pi}{2} - x$
$\Rightarrow \ y = 2\pi - 4x$
On differentiating w.r.t. $x,\$we get
$\Rightarrow \frac{dy}{dx} = \  - 4$

Q. If $y = \sec^{- 1}{\lbrack cosec\ x\rbrack +}{cosec}^{- 1}\left\lbrack \sec x \right\rbrack + \sin^{- 1}{\lbrack\cos{x\rbrack + \cos^{- 1}{\lbrack\sin{x\rbrack,\ \ }}}}\ then\frac{dy}{dx}\ is\ equal$ To

Explanation

Reviewed by Dr. Abubacker

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