If `3tan^(-1)(1/(2+sqrt(3)))-tan^(-1)1/x=tan^(-1)1/3,` then `x` is equal to 1 (b) 2 (c) 3 (d) `sqrt(2)`

Explanation

`3tan^-1(1/(2+sqrt3)) - tan^-1 (1/x) = tan^-1 (1/3)`

`=>3tan^-1(1/(2+sqrt3)**(2-sqrt3)/(2-sqrt3)) = tan^-1 (1/x) + tan^-1 (1/3)`

`=>3tan^-1(2-sqrt3) = tan^-1 (1/x) + tan^-1 (1/3)`

Now, we know, `tan15^@ = 2-sqrt3=>15 = tan^-1(2-sqrt3)`

`:. 3*15^@ = tan^-1((1/3+1/x)/(1-(1/3)(1/x)))`

`=>tan45^@ = (x+3)/(3x-1)`

`=>1(3x-1) = x+3`

`=>2x = 4`

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