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AP EAMCET Engineering 21 Sep 2020 Shift 1 Paper

Section: Mathematics

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Question : 14 of 160

Marks: +1, -0

The derivative of

y=\tan^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]

with respect to

x

is equal to

$-1$
$0$
$\pm 2$
$\pm \frac{1}{2}$

Solution:

It is given that,

y=\tan^{-1}\left[\; \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]

\because \sqrt{1+\sin x}=\sqrt{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}

=\left|\cos\frac{x}{2}+\frac{\sin x}{2}\right|

and

\sqrt{1-\sin x}=\sqrt{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}

=\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|

\therefore \;\; y=\tan^{-1}\left(\cot\frac{x}{2}\right)

when

\cos\frac{x}{2}+\sin\frac{x}{2}

and

\cos\frac{x}{2}-\sin\frac{x}{2}=\frac{\pi}{2}-\frac{x}{2}

are positive

\tan^{-1}\left(\tan\frac{x}{2}\right)=\frac{x}{2}

, when

\cos\frac{x}{2}+\sin\frac{x}{2}

is positive

and

\cos\frac{x}{2}-\sin\frac{x}{2}

is negative

\tan^{-1}\left(\tan\frac{x}{2}\right)=\frac{x}{2}

, when

\cos\frac{x}{2}+\sin\frac{x}{2}

is negative

and

\cos\frac{x}{2}-\sin\frac{x}{2}

is positive

\tan^{-1}\left(\cot\frac{x}{2}\right)=\frac{\pi}{2}-\frac{x}{2}

, when

\cos\frac{x}{2}+\sin\frac{x}{2}

and as

\cos\frac{x}{2}-\sin\frac{x}{2}

is negative.

\text{So},\; \frac{dy}{dx}=\pm \frac{1}{2}

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