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

Correct Answer is : tan⁡2(α2) ²≤ft(α/2)tan2(2α)

Correct Answer is : tan⁡2(α2) ²≤ft(α/2)tan2(2α)

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MHTCET 15 May 2024 Shift 2 Paper

Question : 132 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}(1/y)

and the formula for difference of inverse tangents.

Explanation:

Given

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

Rewrite

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

\tan^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)

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

Using

\tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)

with

a = \frac{1}{\sqrt{\cos \alpha}}

and

b = \sqrt{\cos \alpha}

, we get

ab = 1

Thus

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

Take opposite

= 1-\cos\alpha

and adjacent

= 2\sqrt{\cos\alpha}

for this tangent value.

Then hypotenuse

= \sqrt{(1-\cos\alpha)^2 + (2\sqrt{\cos\alpha})^2} = 1+\cos\alpha

\sin x = \frac{1-\cos\alpha}{1+\cos\alpha}

Using half-angle identities,

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

Answer:

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

, hence option B.

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