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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 119 of 150

Marks: +1, -0

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

\frac{d y}{d x}

x=\frac{\pi}{6}

Solution:

Concept:

Simplify the expression inside

\tan^{-1}

using trigonometric identities and then differentiate.

Explanation:

Given:

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

Use the identities:

1+\sin 2x=(\sin x+\cos x)^2

1-\sin 2x=(\cos x-\sin x)^2

Therefore,

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

x=\frac{\pi}{6}

, we have

\cos x > \sin x

, so the square root simplifies to:

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

Now,

\frac{\sin x+\cos x}{\cos x-\sin x}=\tan\left(\frac{\pi}{4}+x\right)

Thus,

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

x=\frac{\pi}{6}

\frac{\pi}{4}+x=\frac{5\pi}{12}

, which lies in the principal range of

\tan^{-1}

. Hence,

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

Differentiating with respect to

x

, we get:

\frac{dy}{dx}=1

So, the value of

\frac{dy}{dx}

x=\frac{\pi}{6}

1

Answer:

\frac{dy}{dx}=1

Correct option: B

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