Test Index

UPSC NDA 2 2022 Mathematics Solved Paper

© examsnet.com

Question : 48 of 120

Marks: +1, -0

What is the value of tan

\left(\frac{3\pi}{8}\right)

?

$\sqrt{2}$ − 1
$\sqrt{2}$ + 1
1 − $\sqrt{2}$
$-(\sqrt{2} + 1)$

Solution:

Concept:

Use the double angle formula for tangent:

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

.

Explanation:

Let

t = \tan\left(\frac{3\pi}{8}\right)

.

We know that

2 \times \frac{3\pi}{8} = \frac{3\pi}{4}

.

Thus,

\tan\left(\frac{3\pi}{4}\right) = -1

.

Apply the double angle formula:

\tan\left(\frac{3\pi}{4}\right) = \frac{2\tan\left(\frac{3\pi}{8}\right)}{1 - \tan^2\left(\frac{3\pi}{8}\right)}

.

Substitute:

-1 = \frac{2t}{1 - t^2}

.

Multiply both sides by

1 - t^2

:

-1(1 - t^2) = 2t

.

Simplify:

-1 + t^2 = 2t

.

Rearrange:

t^2 - 2t - 1 = 0

.

Solve the quadratic:

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

.

Since

\frac{3\pi}{8}

lies in the first quadrant (between

0

and

\frac{\pi}{2}

), its tangent is positive.

Therefore,

t = 1 + \sqrt{2}

.

Answer:

\sqrt{2} + 1

(Option B).

© examsnet.com

Go to Question:

Prev Question

Next Question