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UPSC NDA I 2024 Mathematics Paper

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Question : 47 of 120

Marks: +1, -0

If

f(\theta)=\frac{1}{1+\tan \theta}

\alpha+\beta=\frac{5 \pi}{4}

, then what is the value to f(α) f(β) ?

$-\frac{1}{2}$
$\frac{1}{2}$
1
2

Solution:

Concept:

Use the tangent addition formula and the given sum to simplify the product of the function values.

Explanation:

Given

f(\theta) = \frac{1}{1 + \tan \theta}

and

\alpha + \beta = \frac{5\pi}{4}

.

Compute

\tan(\alpha + \beta) = \tan\left(\frac{5\pi}{4}\right) = \tan\left(\pi + \frac{\pi}{4}\right) = \tan\frac{\pi}{4} = 1

.

Using

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

, we get:

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

.

Thus,

\tan\alpha + \tan\beta = 1 - \tan\alpha \tan\beta

.

Rearrange:

\tan\alpha + \tan\beta + \tan\alpha \tan\beta = 1

.

Now,

f(\alpha) f(\beta) = \frac{1}{1 + \tan\alpha} \cdot \frac{1}{1 + \tan\beta} = \frac{1}{(1 + \tan\alpha)(1 + \tan\beta)}

.

Expand denominator:

1 + \tan\alpha + \tan\beta + \tan\alpha \tan\beta

.

From above, denominator

= 1 + 1 = 2

.

Therefore,

f(\alpha) f(\beta) = \frac{1}{2}

.

Answer:

\frac{1}{2}

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