If ^{-1}({ α})- ^{-1}({ α})=x, then the value of x is

Correct Answer is : tan⁡2α2 ² α/2tan22α

Correct Answer is : tan⁡2α2 ² α/2tan22α

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 114 of 150

Marks: +1, -0

\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x

\sin x

Solution:

Concept:

Use

\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)

and the standard formula for the difference of inverse tangent functions.

Explanation:

Let

\sqrt{\cos \alpha}

be denoted by

t

. Since

\cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right)

, we rewrite the given equation:

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

Using the identity:

\tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)

Here

A = \frac{1}{t}

and

B = t

. Therefore:

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

So:

\tan x = \frac{1-t^2}{2t} = \frac{1 - \cos \alpha}{2\sqrt{\cos \alpha}}

Now, find

\sin x

using

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

\sqrt{1+\tan^2 x} = \sqrt{1 + \frac{(1-\cos\alpha)^2}{4\cos\alpha}} = \frac{1+\cos\alpha}{2\sqrt{\cos\alpha}}

Hence:

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

Answer:

\sin x = \tan^2\left(\frac{\alpha}{2}\right)

The correct option is D.

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