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Inverse Trigonometric Functions: Find Number Values Satisfying Simultaneously

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

Question
Find the number of values of xx satisfying simultaneously sin⁡−1x=2tan⁡−1x\sin^{-1}x=2\tan^{-1}x and tan⁡−1x(x−1)+csc⁡−11+x−x2=π2\tan^{-1}\sqrt{x(x-1)}+\csc^{-1}\sqrt{1+x-x^{2}}=\dfrac{\pi}{2}.
Solution
Answer: 2
Step 1: For the second equation, rewrite using csc⁡−1u=sin⁡−11u\csc^{-1}u=\sin^{-1}\dfrac{1}{u} or equivalently using the identity tan⁡−1a+csc⁡−1b=π2\tan^{-1}a+\csc^{-1}b=\dfrac{\pi}{2} when a=b2−1a=\sqrt{b^{2}-1}. Equivalently: tan⁡−1x2−x+csc⁡−11−(x2−x)=π2\tan^{-1}\sqrt{x^{2}-x}+\csc^{-1}\sqrt{1-(x^{2}-x)}=\dfrac{\pi}{2}. Step 2: For domain: x2−x\sqrt{x^{2}-x} requires x2−x≥0x^{2}-x\ge 0, and csc⁡−1u\csc^{-1}u requires ∣u∣≥1|u|\ge 1, so 1−(x2−x)≥1\sqrt{1-(x^{2}-x)}\ge 1, i.e., 1−(x2−x)≥11-(x^{2}-x)\ge 1, giving x2−x≤0x^{2}-x\le 0. Step 3: Combining: x2−x=0x^{2}-x=0, so x=0x=0 or x=1x=1. Step 4: Check both in the first equation sin⁡−1x=2tan⁡−1x\sin^{-1}x=2\tan^{-1}x: For x=0x=0: sin⁡−10=0\sin^{-1}0=0 and 2tan⁡−10=02\tan^{-1}0=0. Equal. ✓ For x=1x=1: sin⁡−11=π2\sin^{-1}1=\dfrac{\pi}{2} and 2tan⁡−11=2⋅π4=π22\tan^{-1}1=2\cdot\dfrac{\pi}{4}=\dfrac{\pi}{2}. Equal. ✓ Step 5: Both values satisfy the system. Answer: 22
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