If 2 θ = ² θ - 2, where 0

Correct Answer is : 2−1{2} - 12−1

Correct Answer is : 2−1{2} - 12−1

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UPSC CDS II 2025 Mathematics Paper

Question : 33 of 100

Marks: +1, -0

2 \tan \theta = \sec^{2} \theta - 2

0 < \theta < \frac{\pi}{2}

\cot \theta

equal to?

Solution:

Concept:

Use the identity

\sec^2\theta = 1 + \tan^2\theta

to convert the equation into a quadratic in

\tan\theta

. Then solve for

\tan\theta

and find

\cot\theta

as its reciprocal.

Explanation:

Start with

2\tan\theta = \sec^2\theta - 2

Replace

\sec^2\theta

with

1 + \tan^2\theta

2\tan\theta = (1 + \tan^2\theta) - 2

Simplify:

2\tan\theta = \tan^2\theta - 1

Bring all terms to one side:

\tan^2\theta - 2\tan\theta - 1 = 0

Let

t = \tan\theta

. Then

t^2 - 2t - 1 = 0

Solve using the quadratic formula:

t = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}

Since

0 < \theta < \frac{\pi}{2}

\tan\theta > 0

, so we take

t = 1 + \sqrt{2}

Thus

\tan\theta = 1 + \sqrt{2}

Then

\cot\theta = \frac{1}{\tan\theta} = \frac{1}{1 + \sqrt{2}}

Rationalize: multiply numerator and denominator by

\sqrt{2} - 1

\frac{\sqrt{2} - 1}{(1+\sqrt{2})(\sqrt{2}-1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1

Answer:

Option A:

\sqrt{2} - 1

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