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UPSC NDA I 2022 Mathematics Solved Paper

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Question : 28 of 120

Marks: +1, -0

If

\alpha + \beta = \frac{\pi}{4}

and 2tan α = 1, then what is tan 2β equal to?

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

Solution:

Concept:

Use

\tan(\alpha + \beta)

and double-angle formula for tangent.

Explanation:

Given

2 \tan \alpha = 1

so

\tan \alpha = \frac{1}{2}

.

Also

\alpha + \beta = \frac{\pi}{4}

, so

\beta = \frac{\pi}{4} - \alpha

.

Take tangent:

\tan \beta = \tan\left( \frac{\pi}{4} - \alpha \right)

.

Use

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

:

\tan \beta = \frac{1 - \tan \alpha}{1 + \tan \alpha}

(since

\tan \frac{\pi}{4} = 1

).

Substitute

\tan \alpha = \frac{1}{2}

:

\tan \beta = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{ \frac{1}{2} }{ \frac{3}{2} } = \frac{1}{3}

.

Now find

\tan 2\beta

using

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

:

\tan 2\beta = \frac{2 \cdot \frac{1}{3}}{1 - \left( \frac{1}{3} \right)^2} = \frac{ \frac{2}{3} }{ 1 - \frac{1}{9} } = \frac{ \frac{2}{3} }{ \frac{8}{9} } = \frac{2}{3} \cdot \frac{9}{8} = \frac{18}{24} = \frac{3}{4}

.

Answer:

\frac{3}{4}

, which corresponds to option C.

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