Question

tan inverse(x-1)+tan inverse (x) + tan inverse (x+1)=tan inverse(3x) find x

Answer 1

Solution: Given, tan-1 (x - 1) + tan-1 x + tan-1 (x + 1) = tan-1 (3x)

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

=> tan-1 [{(x - 1) + (x + 1)}/{1 - (x - 1) * (x + 1)}] + tan-1 x = tan-1 (3x)

=> tan-1 [{2x}/{(1 - (x2 - 1)}] + tan-1 x = tan-1 (3x)

=> tan-1 [{2x}/{(1 - (x2 - 1)}] + tan-1 x = tan-1 (3x)

=> tan-1 [{2x}/{(1 - x2 + 1)}] + tan-1 x = tan-1 (3x)

=> tan-1 [{2x}/{(2 - x2 )}] + tan-1 x = tan-1 (3x)

=> tan-1 [{2x/(2 - x2 ) + x}/{1 - 2x2 /{(2 - x2 )}] = tan-1 (3x)

=> tan-1 [{2x + x(2 - x2 )}/(2 - x2 )]/[{1 - 2x2 /{(2 - x2 )}] = tan-1 (3x)

=> tan-1 [{2x + x(2 - x2 )}/{(2 - x2 ) - 2x2 }] = tan-1 (3x)

=> tan-1 [{2x + 2x - x3 )}/{2 - x2 - 2x2 }] = tan-1 (3x)

=> tan-1 [{4x - x3 }/{2 - x2 - 2x2 }] = tan-1 (3x)

=> tan-1 [{4x - x3 }/{2 - 3x2 }] = tan-1 (3x)

=> {4x - x3 }/{2 - 3x2 } = 3x

=> 4x - x3 = {2 - 3x2 ) * 3x

=> 4x - x3 = 6x - 9x3

=> 4x - x3 - 6x + 9x3 = 0

=> 8x3 - 2x = 0

=> 2x(4x2 - 1) = 0

=> 2x*(2x - 1)*(2x + 1) = 0

=> x = 0, 1/2, -1/2

So, the value of x are 0, 1/2, -1/2

Answer 2

Answer:

Step-by-step explanation: