If ^{-1}(1)+ ^{-1}(3)+ ^{-1}(5)+ ^{-1}≤ft(1/4)=π+ ^{-1}≤ft(α/2), then the value of α is...

Correct Answer is : 464146/414146

Correct Answer is : 464146/414146

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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 116 of 150

Marks: +1, -0

\tan ^{-1}(1)+\tan ^{-1}(3)+\tan ^{-1}(5)+\tan ^{-1}\left(\frac{1}{4}\right)=\pi+\tan ^{-1}\left(\frac{\alpha}{2}\right)

\alpha

is...

Solution:

Concept:

The sum of inverse tangents can be evaluated by taking tangent on both sides using the tangent addition formula.

Explanation:

Let

S = \tan^{-1}(1) + \tan^{-1}(3) + \tan^{-1}(5) + \tan^{-1}\left(\frac{1}{4}\right)

First combine

\tan^{-1}(1)

and

\tan^{-1}(3)

using

\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A \tan B}

\tan\left(\tan^{-1}(1)+\tan^{-1}(3)\right)=\frac{1+3}{1-1\cdot 3}=-2.

Now add

\tan^{-1}(5)

\tan\left(\tan^{-1}(1)+\tan^{-1}(3)+\tan^{-1}(5)\right)=\frac{-2+5}{1-(-2)(5)}=\frac{3}{11}.

Now add

\tan^{-1}\left(\frac{1}{4}\right)

\tan S=\frac{\frac{3}{11}+\frac{1}{4}}{1-\frac{3}{11}\cdot \frac{1}{4}}=\frac{\frac{23}{44}}{\frac{41}{44}}=\frac{23}{41}.

Given

S=\pi+\tan^{-1}\left(\frac{\alpha}{2}\right)

. Taking tangent on both sides:

\tan S=\tan\left(\pi+\tan^{-1}\left(\frac{\alpha}{2}\right)\right).

Since

\tan(\pi+\theta)=\tan\theta

, we get

\frac{23}{41}=\frac{\alpha}{2}

Therefore,

\alpha=\frac{46}{41}

Answer:

\alpha=\frac{46}{41}

, which is Option A.

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