If f(α) = {²α - 1}, then what is f(α)+f(β)/1-f(α)f(β) equal to ?

Correct Answer is : f(α+β)f(α + β)f(α+β)

Correct Answer is : f(α+β)f(α + β)f(α+β)

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UPSC NDA 2 2022 Mathematics Solved Paper

Question : 82 of 120

Marks: +1, -0

If f(α) =

\sqrt{\sec^2\alpha - 1}

\frac{f(\alpha)+f(\beta)}{1-f(\alpha)f(\beta)}

equal to ?

Solution:

Concept:

The expression simplifies using

\sqrt{\sec^2\alpha - 1} = \tan\alpha

and the formula

\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

Explanation:

Given

f(\alpha) = \sqrt{\sec^2\alpha - 1}

Using

\sec^2\alpha - 1 = \tan^2\alpha

, we get

f(\alpha) = \sqrt{\tan^2\alpha} = \tan\alpha

Similarly,

f(\beta) = \tan\beta

Now substitute into

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

The right side is exactly

\tan(\alpha+\beta)

Since

f(\theta) = \tan\theta

, we have

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

Answer:

Thus the expression equals

f(\alpha+\beta)

, which corresponds to option B.

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