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UPSC NDA II 2015 Mathematics Paper

Question : 108 of 120

Marks: +1, -0

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

0 < {\rm{x}} < \frac{{\rm{\pi }}}{2}

\frac{{{\rm{dy}}}}{{{\rm{dx}}}}

is equal to

Solution:

Concept:

Use trigonometric identities to simplify the complex argument of

\cot^{-1}

and then differentiate.

Explanation:

For

0 < x < \frac{\pi}{2}

, apply the double-angle formulas.

1+\sin x = (\sin\frac{x}{2} + \cos\frac{x}{2})^2

and

1-\sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2

Since

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

in this interval, the square roots are:

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

and

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

Substitute into the given expression:

y = \cot^{-1}\left[\frac{(\sin\frac{x}{2}+\cos\frac{x}{2}) + (\cos\frac{x}{2}-\sin\frac{x}{2})}{(\sin\frac{x}{2}+\cos\frac{x}{2}) - (\cos\frac{x}{2}-\sin\frac{x}{2})}\right] = \cot^{-1}\left[\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right] = \cot^{-1}\left(\cot\frac{x}{2}\right)

Since

\frac{x}{2} \in (0, \frac{\pi}{4})

, which lies within the principal range of

\cot^{-1}

, we get

y = \frac{x}{2}

Differentiate with respect to

x

\frac{dy}{dx} = \frac{1}{2}

Answer:

\frac{1}{2}

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