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UPSC CDS II 2024 Math Paper

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Question : 16 of 100

Marks: +1, -0

If

64^{\sin^{2}\theta}+64^{\cos^{2}\theta}=16

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

, then what is the value of tanθ + cotθ?

1
2
3
4

Solution:

Concept:

Use substitution and the identity

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

to solve the exponential equation.

Explanation:

Let

x = 64^{\sin^2\theta}

.

Then

64^{\cos^2\theta} = 64^{1 - \sin^2\theta} = \frac{64}{x}

.

The given equation becomes

x + \frac{64}{x} = 16

.

Multiply both sides by

x

:

x^2 + 64 = 16x

.

Rearrange:

x^2 - 16x + 64 = 0

.

Factor:

(x-8)^2 = 0

, so

x = 8

.

Thus

64^{\sin^2\theta} = 8 = 64^{\frac{1}{2}}

.

Since bases are equal,

\sin^2\theta = \frac{1}{2}

.

Hence

\cos^2\theta = 1 - \frac{1}{2} = \frac{1}{2}

.

Taking square root (positive as

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

):

\sin\theta = \cos\theta = \frac{1}{\sqrt{2}}

.

Therefore

\theta = 45^{\circ}

(or

\frac{\pi}{4}

).

Then

\tan\theta = \cot\theta = 1

.

So

\tan\theta + \cot\theta = 1 + 1 = 2

.

Answer:

The value is

2

, which corresponds to option B.

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