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UPSC CDS II 2024 Math Paper

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Question : 14 of 100

Marks: +1, -0

If

0⩽(\alpha -\beta )⩽(\alpha +\beta )⩽\frac{\pi }{2}tan(\alpha +\beta )=\sqrt{3}

tan(\alpha -\beta )=\frac{1}{\sqrt{3}}

, then what is tanα.cot2β equal to?

1
√2
√3
$\frac{1}{\sqrt{3}}$

Solution:

Concept:

Use the given tangent values to find

\alpha

and

\beta

from inverse trigonometric angles, then compute

\tan\alpha \cdot \cot 2\beta

.

Explanation:

Given:

\tan(\alpha + \beta) = \sqrt{3}

and

\tan(\alpha - \beta) = \frac{1}{\sqrt{3}}

.

Since

0 \le (\alpha - \beta) \le (\alpha + \beta) \le \frac{\pi}{2}

, we know

\alpha + \beta = \frac{\pi}{3}

and

\alpha - \beta = \frac{\pi}{6}

.

Add the two equations:

(\alpha+\beta)+(\alpha-\beta)=\frac{\pi}{3}+\frac{\pi}{6}

gives

2\alpha = \frac{\pi}{2}

, so

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

.

Substitute

\alpha

into

\alpha+\beta=\frac{\pi}{3}

:

\frac{\pi}{4}+\beta=\frac{\pi}{3}

gives

\beta = \frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}

.

Now,

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

.

Compute

2\beta = 2 \times \frac{\pi}{12} = \frac{\pi}{6}

. Then

\tan(2\beta) = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}

, so

\cot(2\beta) = \frac{1}{\tan(2\beta)} = \sqrt{3}

.

Finally,

\tan\alpha \cdot \cot 2\beta = 1 \times \sqrt{3} = \sqrt{3}

.

Answer:

\sqrt{3}

(Option C).

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