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Find the value of x `tan^-1 ((x-1)/(x+1)) + tan^-1 ((2x-1)/(2x+1)) = tan^-1 (23/36)` |
Answer» We know, `tan^(-1)x+tan^(-1)y = tan^-1((x+y)/(1-xy))` ` tan^(-1)((x-1)/(x+1))+tan^(-1)((2x-1)/(2x+1))= tan^(-1)((x-1)/(x+1)+(2x-1)/(2x+1))/((1-((x-1)/(x+1))((2x-1)/(2x+1)))` `= tan^(-1)((((x-1)(2x+1)+(2x-1)(x+1))/((x+1)(2x+1)))/(((x+1)(2x+1)-(x-1)(2x-1))/((x+1)(2x+1))))` `=tan^(-1)(((2x^2-x-1)+(2x^2+x-1))/((2x^2+3x+1) - (2x^2-3x+1)))` `=tan^(-1)((4x^2-2)/(6x))` So, `tan^(-1)((4x^2-2)/(6x)) = tan^(-1)(23/36)` `:. (4x^2-2)/(6x) = 23/36` `=> 24x^2 -12 = 23x` `=>24x^2-23x-12 = 0` `=>24x^2-32x+9x-12 = 0` `=>8x(3x-4)+3(3x-4) = 0` `=>(3x-4)(8x+3) = 0` `:. x = 4/3 and x = -8/3` |
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