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UPSC CDS II 2025 Mathematics Paper

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Question Numbers: 45-46

Let

\frac{sin\alpha }{sin\beta }=\frac{4\sqrt{2}}{3}

and

\frac{cos\alpha }{cos\beta }=\frac{2\sqrt{3}}{9}

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Question : 46 of 100

Marks: +1, -0

What is

\tan^{2}\beta

equal to? .

1/√2
3/√2
$\frac{1}{3}$
$\frac{2}{3}$

Solution:

Concept:

Use the identity

\sin^2\alpha + \cos^2\alpha = 1

and the given ratios to find

\sin\beta

and

\cos\beta

, then compute

\tan^2\beta

.

Explanation:

Let

\sin\alpha = \frac{4\sqrt{2}}{3} \sin\beta

and

\cos\alpha = \frac{2\sqrt{3}}{9} \cos\beta

.

Square and add:

\left(\frac{4\sqrt{2}}{3}\right)^2 \sin^2\beta + \left(\frac{2\sqrt{3}}{9}\right)^2 \cos^2\beta = 1

.

This gives

\frac{32}{9} \sin^2\beta + \frac{12}{81} \cos^2\beta = 1

. Simplify to

\frac{32}{9} \sin^2\beta + \frac{4}{27} \cos^2\beta = 1

.

Multiply by 27:

96 \sin^2\beta + 4 \cos^2\beta = 27

.

Replace

\cos^2\beta = 1 - \sin^2\beta

:

96 \sin^2\beta + 4(1 - \sin^2\beta) = 27

.

So

96 \sin^2\beta + 4 - 4 \sin^2\beta = 27

, hence

92 \sin^2\beta = 23

.

Thus

\sin^2\beta = \frac{1}{4}

, so

\sin\beta = \frac{1}{2}

(positive).

Then

\cos^2\beta = 1 - \frac{1}{4} = \frac{3}{4}

, giving

\cos\beta = \frac{\sqrt{3}}{2}

.

Therefore

\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}

.

Finally,

\tan^2\beta = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}

.

Answer:

C.

\frac{1}{3}

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