Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If x=sin⁡−1(a6+1)+cos⁡−1(a4+1)−tan⁡−1(a2+1)x=\sin^{-1}(a^{6}+1)+\cos^{-1}(a^{4}+1)-\tan^{-1}(a^{2}+1), a∈Ra\in\mathbb{R}, then find the value of sec⁡2x\sec^{2}x.
Solution
Answer: 2
Step 1: Since sin⁡−1u\sin^{-1}u is defined only for u∈[−1,1]u\in[-1,1], we need a6+1∈[−1,1]a^{6}+1\in[-1,1]. Step 2: As a6≥0a^{6}\ge 0, we have a6+1≥1a^{6}+1\ge 1. Combined with a6+1≤1a^{6}+1\le 1: a6=0a^{6}=0, so a=0a=0. Step 3: Substituting a=0a=0:
x=sin⁡−11+cos⁡−11−tan⁡−11=π2+0−π4=π4x=\sin^{-1}1+\cos^{-1}1-\tan^{-1}1=\dfrac{\pi}{2}+0-\dfrac{\pi}{4}=\dfrac{\pi}{4}
 Step 4: Therefore:
sec⁡2x=sec⁡2π4=(2)2=2\sec^{2}x=\sec^{2}\dfrac{\pi}{4}=(\sqrt{2})^{2}=2
 Answer: 22
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