Solve the following equation for `x` : `3sin^(-1)(2x)/(1+x^2)-4cos^(-1)(1-x^2)/(1+x^2)+2tan^(-1)(2x)/(1-x^2)=pi/3`

Explanation

Given,

`3sin^(−1)[(2x)/(1+x^2)]−4cos^(−1)(1−x^2)/([1+x^2)​]+2tan^(−)1[(2x)/(1−x^2)​]=(pi)/3​`



`2tan^(−1)x=tan^(−1)[(2x)/(1−x^2)​]=sin^(−1)[(2x)/(1+x^2)​]=cos^(−1)[(1−x^2)/(1+x^2)​] `

∴ `3[2tan^(−1)x]−4[2tan^(−1)x]+2[2tan^(−1)x]=(pi)/3`

`=>6tan^(−1)x−8tan^(−1)x+4tan^(−1)x=(pi)/3​`

`=>2tan^(−1)x=(pi)/3​`

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