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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
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B
13
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C
3
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D
34
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E
32
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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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