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Mathematics Question on Inverse Trigonometric Functions

If 3 sin⁡−1(2x1+x2)−4 cos⁡−1(1−x21+x2)+2 tan−1(2x1−x2)=π33\,\sin^{-1}\left(\frac{2x}{1+x^2}\right)-4\,\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2\,tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}, then xx is equal to

A

13\frac{1}{\sqrt3}

B

−13-\frac{1}{\sqrt3}

C

3\sqrt3

D

−34-\frac{\sqrt3}{4}

Answer

13\frac{1}{\sqrt3}

Explanation

Solution

Put x=tan θx=tan\,\theta ∴3 sin−12 tan θ1+tan2θ−4⋅cos−11−tan2θ1+tan2θ\therefore 3\,sin^{-1}\frac{2\,tan\,\theta}{1+tan^{2}\theta} - 4\cdot cos^{-1} \frac{1-tan^{2} \theta}{1+tan^{2}\theta} +2 tan−12 tan θ1−tan2θ=π3+2\,tan^{-1} \frac{2\,tan\,\theta}{1-tan^{2} \theta} = \frac{\pi}{3} ⇒3 sin−1(sin 2 θ)−4 cos−1(cos 2 θ)+2 sin−1(tan 2 θ)=π3\Rightarrow 3\,sin^{-1} \left(sin\,2\,\theta\right) -4\,cos^{-1}\left(cos\,2\,\theta\right)+2\,sin^{-1} \left(tan\, 2\,\theta\right) = \frac{\pi}{3} ⇒3(2 θ)−4(2 θ)+2(2 θ)=π3\Rightarrow 3\left(2\,\theta\right)-4\left(2\,\theta\right)+2\left(2\,\theta\right) = \frac{\pi}{3} ⇒2 θ=π3\Rightarrow 2\,\theta= \frac{\pi}{3} ⇒θ=π6\Rightarrow \theta = \frac{\pi}{6} ⇒x=tanπ6\Rightarrow x=tan \frac{\pi}{6} =13= \frac{1}{\sqrt{3}}