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UPSC NDA II 2025 Mathematics Paper

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Question : 46 of 120

Marks: +1, -0

What is

cot^{-1}9+ cosec -1(\frac{\sqrt{41}}{4})

equal to?

$\frac{\pi }{4}$
$\frac{\pi }{3}$
$\frac{\pi }{2}$
$\pi$

Solution:

Concept:

We use the identities

cot^{-1}x = \tan^{-1}(1/x)

and

csc^{-1}x = \tan^{-1}(\frac{1}{\sqrt{x^2-1}})

or construct a right triangle. Then apply

\tan^{-1}a + \tan^{-1}b = \tan^{-1}(\frac{a+b}{1-ab})

when

ab<1

.

Explanation:

First,

cot^{-1}9 = \tan^{-1}(\frac{1}{9})

.

For

csc^{-1}(\frac{\sqrt{41}}{4})

, let

\theta = csc^{-1}(\frac{\sqrt{41}}{4})

, so

csc\theta = \frac{\sqrt{41}}{4} = \frac{\text{hypotenuse}}{\text{opposite}}

. Thus opposite = 4, hypotenuse =

\sqrt{41}

. Adjacent side =

\sqrt{(\sqrt{41})^2 - 4^2} = \sqrt{41-16}=5

. Hence

\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{5}

, so

csc^{-1}(\frac{\sqrt{41}}{4}) = \tan^{-1}(\frac{4}{5})

.

Now sum:

\tan^{-1}(\frac{1}{9}) + \tan^{-1}(\frac{4}{5})

. Since

\frac{1}{9}\cdot\frac{4}{5} = \frac{4}{45} < 1

, we use identity:

\tan^{-1}a + \tan^{-1}b = \tan^{-1}(\frac{a+b}{1-ab})

. Here

a=\frac{1}{9}, b=\frac{4}{5}

.

Compute

\frac{a+b}{1-ab} = \frac{\frac{1}{9}+\frac{4}{5}}{1 - \frac{1}{9}\cdot\frac{4}{5}} = \frac{\frac{5+36}{45}}{1-\frac{4}{45}} = \frac{\frac{41}{45}}{\frac{41}{45}} = 1

. So the sum equals

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

.

Answer:

\frac{\pi}{4}

(Option A)

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