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MHTCET 25 Apr 2025 Shift 2 Paper

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Question : 120 of 150

Marks: +1, -0

If

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

x^{2}+1=

-1
2
1
-2

Solution:

Concept:

Use the identity

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

and solve the resulting quadratic equation.

Explanation:

Given:

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

Let

y = \tan^{-1} x

. Then

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

.

Substitute into the equation:

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

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

2y^2 - \pi y + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0

2y^2 - \pi y - \frac{3\pi^2}{8} = 0

Multiply by

8

:

16y^2 - 8\pi y - 3\pi^2 = 0

Solving:

y = \frac{8\pi \pm 16\pi}{32}

So

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

or

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

Since

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

, we take

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

Thus

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

Therefore

x^2 + 1 = (-1)^2 + 1 = 2

Answer:

Option B:

2

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