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Mathematics · Inverse Trigonometric Functions

JEE Advanced 2022 — Paper 1 — Question 1

Considering only the principal values of the inverse trigonometric functions, the value of

32cos⁡−122+π2+14sin⁡−122π2+π2+tan⁡−12π\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^{2}}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^{2}}+\tan ^{-1} \frac{\sqrt{2}}{\pi} is \qquad

Answer: 2.35

Numerical answer — enter this value.

Step-by-step solution

Let tan⁡θ=π2⇒π4≤θ≤π2⇒π=2tan⁡θ\tan \theta=\frac{\pi}{\sqrt{2}} \Rightarrow \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2} \Rightarrow \pi=\sqrt{2} \tan \theta

32cos⁡−122+2tan⁡2θ+14sin⁡−1(222tan⁡θ2+2tan⁡2θ)+tan⁡−1(22tan⁡θ)\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+2 \tan ^{2} \theta}}+\frac{1}{4} \sin ^{-1}\left(\frac{2 \sqrt{2} \sqrt{2} \tan \theta}{2+2 \tan ^{2} \theta}\right)+\tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{2} \tan \theta}\right)

=32cos⁡−1(cos⁡θ)+14sin⁡−1(sin⁡2θ)+tan⁡−1(cot⁡θ)=\frac{3}{2} \cos ^{-1}(\cos \theta)+\frac{1}{4} \sin ^{-1}(\sin 2 \theta)+\tan ^{-1}(\cot \theta)

32θ+14(π−2θ)+π2−θ=π4+π2=3π4\frac{3}{2} \theta+\frac{1}{4}(\pi-2 \theta)+\frac{\pi}{2}-\theta=\frac{\pi}{4}+\frac{\pi}{2}=\frac{3 \pi}{4}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Fundamentals of the six ITFs