If 3 ^{-1}≤ft({2 x}{1+x²})-4 ^{-1}≤ft({1-x²}{1+x²})+2 ^{-1}≤ft({2 x}{1-x²})=π/3 then the value of x=

Correct Answer is : 13{1}{{3}}31

Correct Answer is : 13{1}{{3}}31

Test Index

MHTCET 22 Apr 2025 Shift 1 Paper

Question : 146 of 150

Marks: +1, -0

3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}

x=

Solution:

Concept:

Use the substitution

x = \tan\theta

to convert the inverse trigonometric terms into standard angle forms.

Explanation:

Let

x = \tan\theta

, so

\theta = \tan^{-1}x

Substitute into the given equation:

3 \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) - 4 \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) + 2 \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) = \frac{\pi}{3}

Using standard identities:

\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta

\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta

\frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta

Therefore, the equation becomes:

3\sin^{-1}(\sin 2\theta) - 4\cos^{-1}(\cos 2\theta) + 2\tan^{-1}(\tan 2\theta) = \frac{\pi}{3}

For principal values,

3(2\theta) - 4(2\theta) + 2(2\theta) = \frac{\pi}{3}

Simplifying:

6\theta - 8\theta + 4\theta = 2\theta = \frac{\pi}{3}

\theta = \frac{\pi}{6}

Since

\tan^{-1}x = \theta

, we get

x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}

Answer:

The value of

x

\frac{1}{\sqrt{3}}

, i.e., Option C.

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