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UPSC NDA II 2024 Mathematics Paper

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Question : 22 of 120

Marks: +1, -0

What is

\cot^{2}(\sec^{-1} 2) + \tan^{2}(\csc^{-1} 3)

equal to ?

$\frac{11}{12}$
$\frac{11}{24}$
$\frac{7}{24}$
$\frac{1}{24}$

Solution:

Concept:

Evaluate inverse trigonometric functions and then compute squares of trigonometric ratios.

Explanation:

Let

\theta = \sec^{-1} 2

.

Then

\sec \theta = 2

, so

\cos \theta = \frac{1}{2}

.

Thus

\theta = 60^{\circ} = \frac{\pi}{3}

.

Therefore

\cot \theta = \cot 60^{\circ} = \frac{1}{\sqrt{3}}

.

So

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

.

Let

\phi = \csc^{-1} 3

.

Then

\csc \phi = 3

, so

\sin \phi = \frac{1}{3}

.

Using

\cos^{2}\phi = 1 - \sin^{2}\phi = 1 - \frac{1}{9} = \frac{8}{9}

, we get

\cos \phi = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}

.

Hence

\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{8}}

.

So

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

.

Now add:

\frac{1}{3} + \frac{1}{8} = \frac{8+3}{24} = \frac{11}{24}

.

Answer:

\frac{11}{24}

(Option B).

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