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Mathematics · Inverse Trigonometric Functions

JEE Main 2026 — 21 January, Evening Shift — Question 23

Let the maximum value of (sin⁡−1x)2+(cos⁡−1x)2\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2} for x∈[−32,12]x \in\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right] be mnπ2\frac{m}{n} \pi^{2}, where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1. Then m+n\mathrm{m}+\mathrm{n} is equal to ____\_\_\_\_ .

Answer: 65

Numerical answer — enter this value.

Step-by-step solution

(sin⁡−1x)2+(cos⁡−1x)2\left(\sin ^{-1} \mathrm{x}\right)^{2}+\left(\cos ^{-1} \mathrm{x}\right)^{2}

=(sin⁡−1x+cos⁡−1x)2−2sin⁡−1xcos⁡−1x=\left(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}\right)^{2}-2 \sin ^{-1} \mathrm{x} \cos ^{-1} \mathrm{x}

=π24−2(sin⁡−1x)(π2−sin⁡−1x)=\frac{\pi^{2}}{4}-2\left(\sin ^{-1} \mathrm{x}\right)\left(\frac{\pi}{2}-\sin ^{-1} \mathrm{x}\right)

=2(sin⁡−1x−π4)2+π28=2\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8} \quad

where sin⁡−1x∈[−π3,π4]\sin ^{-1} \mathrm{x} \in\left[\frac{-\pi}{3}, \frac{\pi}{4}\right]

Then max value occurs at sin⁡−1x=−π3\sin ^{-1} \mathrm{x}=\frac{-\pi}{3}

Which is 2(π3+π4)2+π28=29π2362\left(\frac{\pi}{3}+\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=\frac{29 \pi^{2}}{36}

⇒m=29\Rightarrow \mathrm{m}=29 and n=36\mathrm{n}=36

∴m+n=65\therefore \mathrm{m}+\mathrm{n}=65

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Equations, Inequations and Identitites involving ITFs