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UPSC NDA 2 2022 Mathematics Solved Paper

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Question : 49 of 120

Marks: +1, -0

\text{What is } \tan^{-1} \cot(\csc^{-1} 2) \text{ equal to ?}

$\frac{\pi}{8}$
$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$

Solution:

Concept:

Use the identity

1 + \cot^2 x = \csc^2 x

and the principal range of

\csc^{-1} x

.

Explanation:

Let

y = \csc^{-1} 2

.

Then

y \in (0, \pi/2]

and

\csc y = 2

.

Using

\cot^2 y = \csc^2 y - 1

, we get

\cot^2 y = 4 - 1 = 3

.

Hence

\cot y = \sqrt{3}

(positive because

y

is in the first quadrant).

Now

\tan^{-1}(\cot(\csc^{-1}2)) = \tan^{-1}(\cot y) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

.

The value

\frac{\pi}{3}

lies in the principal range of

\tan^{-1}

, so it is acceptable.

Answer:

\frac{\pi}{3}

(option D).

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