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MHTCET 16 May 2024 Shift 2 Paper

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Question : 102 of 150

Marks: +1, -0

If

\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\frac{1}{2} \cos ^{-1} x

x

is

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{4}{5}$

Solution:

Concept:

Use the addition formula for inverse tangent and the identity linking

2\tan^{-1}t

with

\cos^{-1}

.

Explanation:

Given:

\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{2}{9}\right)=\frac{1}{2}\cos^{-1}x

Apply the formula:

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

Here,

a=\frac{1}{4}

and

b=\frac{2}{9}

.

\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\cdot\frac{2}{9}}\right)

= \tan^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right) = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\left(\frac{1}{2}\right)

So,

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

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

Use the identity

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

with

t=\frac{1}{2}

.

\cos^{-1}x = \cos^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}\right) = \cos^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right) = \cos^{-1}\left(\frac{3}{5}\right)

Therefore,

x=\frac{3}{5}

.

Answer:

x=\frac{3}{5}

, i.e. option C.

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