Question

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

Answer 1

Hello,

$sin(a-b)=sin(a)cos(b)-cos(a)sin(b)\\cos(a-b)=cos(a)cos(b)+sin(a)sin(b)\\tan(a-b)=\dfrac{sin(a-b)}{cos(a-b)} =\dfrac{sin(a)cos(b)-cos(a)sin(b)}{cos(a)cos(b)(1+\dfrac{sin(a)sin(b)}{cos(a)cos(b)}} \\\\\boxed{tan(a-b)=\dfrac{tan(a)-tan(b)}{1-tan(a)*tan(b)}}\\\\ Let\ u=tan(a)\ \Rightarrow\ a=arctan(u)\\and\ v=tan(b)\ \Rightarrow\ b=arctan(v)\\arctan(tan(a-b))=a-b=arctan(u)-arctan(v)\\tan(a)-tan(b)=u-v\\1+tan(a)tan(b)=1+uv\\arctan(tan(a-b))=a-b=arctan(u)-arctan(v)=arctan(\dfrac{u-v}{1+uv})$

$arctan(\dfrac{6x-8x^3}{1-12x^2})-arctan(\dfrac{4x}{1-4x^2} )\\=arctan(\dfrac{\dfrac{6x-8x^3}{1-12x^2}-\dfrac{4x}{1-4x^2}}{1+\dfrac{6x-8x^3}{1-12x^2}*\dfrac{4x}{1-4x^2}} ) \\\\=arctan(\dfrac{32x^5+16x^3+2x}{16x^4+8x^2+1})=arctan(2x)$