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Questions
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.
Solution
`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=1-6x^2`
`6x^2+5x-1=0 `
`6x^2+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`
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