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UPSC NDA I 2024 Mathematics Paper

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Question : 42 of 120

Marks: +1, -0

What is

\sqrt{15+\cot^2\left(\frac{\pi}{4}-2\cot^{-1}3\right)}

equal to ?

1
7
8
16

Solution:

Concept:

Use inverse trigonometric identities to simplify the given expression.

Explanation:

First, rewrite

2\cot^{-1}3

using

\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)

:

2\cot^{-1}3 = 2\tan^{-1}\left(\frac{1}{3}\right)

.

Apply the formula

2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)

:

2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2\cdot\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) = \tan^{-1}\left(\frac{2}{3}\times\frac{9}{8}\right) = \tan^{-1}\frac{3}{4}

.

Now the original expression becomes:

\sqrt{15+\cot^2\left(\frac{\pi}{4} - \tan^{-1}\frac{3}{4}\right)}

.

Since

\frac{\pi}{4} = \tan^{-1}1

, we have:

\sqrt{15+\cot^2\left(\tan^{-1}1 - \tan^{-1}\frac{3}{4}\right)}

.

Use

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)

:

\tan^{-1}1 - \tan^{-1}\frac{3}{4} = \tan^{-1}\left(\frac{1-\frac{3}{4}}{1+1\cdot\frac{3}{4}}\right) = \tan^{-1}\left(\frac{\frac{1}{4}}{\frac{7}{4}}\right) = \tan^{-1}\frac{1}{7}

.

Thus expression =

\sqrt{15+\cot^2\left(\tan^{-1}\frac{1}{7}\right)}

.

But

\tan^{-1}\frac{1}{7} = \cot^{-1}7

(since

\cot^{-1}x = \tan^{-1}\frac{1}{x}

), so:

\sqrt{15+\cot^2\left(\cot^{-1}7\right)} = \sqrt{15+7^2} = \sqrt{15+49} = \sqrt{64} = 8

.

Answer:

8

(Option C).

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