Question

If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.

If `tan^-1 (2x) + tan^-1(3x) = pi/4` then find the value of x, where 0 < 3x < 1.

Answer 1

`tan^-1(2x)+tan^-1(3x)=pi/4`

`tan^-1((2x+3x)/(1-(2x)(3x)))=pi/4`

`therefore (5x)/(1-6x^2)=tan(pi/4)`

`(5x)/(1-6x^2)=1`

5x = 1 − 6x²

6x² + 5x − 1 = 0

6x² + 6x − x − 1 = 0

6x(x + 1) − 1(x + 1) = 0

(x + 1)(6x − 1) = 0

x = −1 or x = `1/6`

But x = −1 does not satisfy ` tan^-1(2x)+tan^-1(3x)=pi/4`

`x=1/6`