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MHTCET 11 May 2024 Shift 1 Paper

Question : 126 of 150

Marks: +1, -0

The value of

\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x

Solution:

Concept:

Use the substitution

x = \cos 2\theta

to simplify the inverse tangent expression and then combine like terms.

Explanation:

Let

x = \cos 2\theta

, so

\theta = \frac{1}{2}\cos^{-1} x

The tangent term becomes:

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

Using identities:

\sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta

and

\sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta

Thus the expression becomes:

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

Now,

\frac{1-\tan\theta}{1+\tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)

, so:

\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right) = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x

Adding

\frac{1}{2}\cos^{-1} x

\left(\frac{\pi}{4} - \frac{1}{2}\cos^{-1} x\right) + \frac{1}{2}\cos^{-1} x = \frac{\pi}{4}

Answer:

The value is

\frac{\pi}{4}

, which corresponds to option B.

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