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MHTCET 16 Apr 2026 Shift 2 Paper

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Question : 117 of 150

Marks: +1, -0

If

\left(\tan ^{-1} x\right)^{2}+\left(\cot ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8}

, then the value of

x

is equal to...

-1
-2
1
2

Solution:

Concept:

Use the identity

\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}

and the given quadratic condition to find

x

.

Explanation:

Let

A = \tan^{-1} x

and

B = \cot^{-1} x

.

Then

A + B = \frac{\pi}{2}

.

Given

A^2 + B^2 = \frac{5\pi^2}{8}

.

Using

(A+B)^2 = A^2 + B^2 + 2AB

, we get:

\left(\frac{\pi}{2}\right)^2 = \frac{5\pi^2}{8} + 2AB

\frac{\pi^2}{4} = \frac{5\pi^2}{8} + 2AB

2AB = \frac{\pi^2}{4} - \frac{5\pi^2}{8} = -\frac{3\pi^2}{8}

So

AB = -\frac{3\pi^2}{16}

.

Now

A

and

B

have sum

\frac{\pi}{2}

and product

-\frac{3\pi^2}{16}

.

Thus the two values are

A = -\frac{\pi}{4}

and

B = \frac{3\pi}{4}

.

Since

\tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, we take

A = -\frac{\pi}{4}

.

Therefore,

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

.

Answer:

x = -1

, which is option A.

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