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Question: \(y=\tan ^{-1}\left(\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right)\) express in terms of sin inver...

y=tan⁡−1(x−1−x2x+1−x2)y=\tan ^{-1}\left(\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right) express in terms of sin inverse x

A

sin⁡−1x−π4\sin ^{-1} x-\frac{\pi}{4}

B

sin⁡−1x+π4\sin ^{-1} x+\frac{\pi}{4}

C

π4−sin⁡−1x\frac{\pi}{4}-\sin ^{-1} x

D

3π4+sin⁡−1x\frac{3\pi}{4}+\sin ^{-1} x

Answer

sin⁡−1x−π4\sin ^{-1} x-\frac{\pi}{4}

Explanation

Solution

Let x=cos⁡ϕx = \cos \phi. Then 1−x2=sin⁡ϕ\sqrt{1-x^2} = \sin \phi for ϕ∈[0,π]\phi \in [0, \pi]. The expression inside the tan⁡−1\tan^{-1} becomes cos⁡ϕ−sin⁡ϕcos⁡ϕ+sin⁡ϕ=1−tan⁡ϕ1+tan⁡ϕ=tan⁡(π4−ϕ)\frac{\cos \phi - \sin \phi}{\cos \phi + \sin \phi} = \frac{1 - \tan \phi}{1 + \tan \phi} = \tan(\frac{\pi}{4} - \phi). So, y=tan⁡−1(tan⁡(π4−ϕ))y = \tan^{-1}(\tan(\frac{\pi}{4} - \phi)). For the principal value, we need −π2<π4−ϕ<π2-\frac{\pi}{2} < \frac{\pi}{4} - \phi < \frac{\pi}{2}. This implies −3π4<−ϕ<π4-\frac{3\pi}{4} < -\phi < \frac{\pi}{4}, or −π4<ϕ<3π4-\frac{\pi}{4} < \phi < \frac{3\pi}{4}. Since ϕ=cos⁡−1x\phi = \cos^{-1} x, the range of ϕ\phi is [0,π][0, \pi]. Thus, for ϕ∈[0,3π4)\phi \in [0, \frac{3\pi}{4}), which corresponds to x∈(−12,1]x \in (-\frac{1}{\sqrt{2}}, 1], we have y=π4−ϕy = \frac{\pi}{4} - \phi. Using ϕ=cos⁡−1x=π2−sin⁡−1x\phi = \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x, we get y=π4−(π2−sin⁡−1x)=sin⁡−1x−π4y = \frac{\pi}{4} - (\frac{\pi}{2} - \sin^{-1} x) = \sin^{-1} x - \frac{\pi}{4}.