Question

# If $$\displaystyle 3\:sin^{-1}\left(\frac{2x}{1+x^{2}}\right)-4\:cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2tan^{-1}\left(\frac{2x}{1-x^{2}}\right)=\frac{\pi}{3}$$, where $$|x|< 1$$, then $$x$$ is equal to

A
13
B
13
C
3
D
34
E
32

Solution

## The correct option is A $$\displaystyle \frac{1}{\sqrt{3}}$$Since $$|x|<1$$Substituting $$x=tan\theta$$The above expression reduces to$$3sin^{-1}(sin2\theta)-4cos^{-1}(cos2\theta)+2tan^{-1}(tan2\theta)$$$$=6\theta-8\theta+4\theta$$$$=2\theta$$$$=\dfrac{\pi}{3}$$Therefore$$\theta=\frac{\pi}{6}$$$$tan^{-1}{x}=\frac{\pi}{6}$$$$x=tan(\frac{\pi}{6})$$$$x=\frac{1}{\sqrt{3}}$$Mathematics

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