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NCERT Exemplar · Q34

Q.If ∣x∣≤1|x|\le1, then 2tan⁡−1x+sin⁡−1(2x1+x2)2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right) is equal to
(A) 4tan⁡−1x4\tan^{-1}x
(B) 00
(C) π2\frac{\pi}{2}
(D) π\pi

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On ∣x∣≤1|x|\le1 the identity sin⁡−1 ⁣(2x1+x2)=2tan⁡−1x\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)=2\tan^{-1}x holds, so the expression is 2tan⁡−1x+2tan⁡−1x=4tan⁡−1x2\tan^{-1}x+2\tan^{-1}x=4\tan^{-1}x — option (A).

The idea

The fraction 2x1+x2\frac{2x}{1+x^2} is exactly sin⁡2θ\sin2\theta when x=tan⁡θx=\tan\theta. The only subtlety is whether sin⁡−1(sin⁡2θ)=2θ\sin^{-1}(\sin2\theta)=2\theta, which needs 2θ2\theta inside the sin⁡−1\sin^{-1} range [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Step 1 — Substitute

Let x=tan⁡θx=\tan\theta with θ=tan⁡−1x\theta=\tan^{-1}x. Since ∣x∣≤1|x|\le1, θ∈[−π4,π4]\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]. Then

2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡2θ.\frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta.

Step 2 — Peel off the sin⁡−1\sin^{-1}

Because θ∈[−π4,π4]\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right], we have 2θ∈[−π2,π2]2\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right], the principal range of sin⁡−1\sin^{-1}. Hence …

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