Question

 

 

 

Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`

 

 

 

Answer 1

 

 

 

Consider the left hand side

L.H.S = `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))`

We know that, 

`tan^(-1)(A)-tan^(-1)(B)= tan^(-1)((A-B)/(1+AB))`

Thus, L.H.S = `tan^(-1)(((6x-8x^3)/(1-12x^2)-(4x)/(1-4x^2))/(1+((6x-8x^3)/(1-12x^2))((4x)/(1-4x^2))))`

`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/(((1-12x^2)(1-4x^2))/(1+(4x(6x-8x^3))/((1-12x^2)(1-4x^2)))))`

`=tan^(-1)((((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)))/(((1-12x^2)(1-4x^2)+4x(6x-8x^3))/((1-12x^2)(1-4x^2))))`

`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)+4x(6x-8x^3)))`

`=tan^(-1)((6x-24x^3-8x^3+32x^5-4x+48x^3)/(1-4x^2-12x^2+48x^4+24x^2-32x^4))`

 `=tan^(-1)((32x^5+16x^3+2x)/(16x^4+8x^2+1))`

 `=tan^(-1)((2x(16x^4+8x^2+1))/(16x^4+8x^2+1))`

 = tan^-12x

 Thus, L.H.S=R.H.S