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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 149 of 150

Marks: +1, -0

y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), 0 \leqslant x < \frac{\pi}{2}

y^{\prime}\left(\frac{\pi}{6}\right)=

Solution:

Concept:

Simplify the given inverse trigonometric expression using half-angle identities before differentiating.

Explanation:

For

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

, both

\cos\frac{x}{2}

and

\sin\frac{x}{2}

are positive.

Use the identities:

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

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

Thus,

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

Divide numerator and denominator by

\cos\frac{x}{2}

y=\tan^{-1}\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right)=\tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]

Since

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

, the angle

\frac{\pi}{4}+\frac{x}{2}

lies within the principal range, so:

y=\frac{\pi}{4}+\frac{x}{2}

Differentiate with respect to

x

y'=\frac{1}{2}

Therefore, at

x=\frac{\pi}{6}

y'\left(\frac{\pi}{6}\right)=\frac{1}{2}

Answer:

\frac{1}{2}

, which is option D.

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