Question 737716
Rewrite the expression as an algebraic expression in x. Assume x>0
{{{cos(2tan^-1(1/x))}}}
let O=opposite side
let A=adjacent side
H=hypotenuse
...
let y=tan^-1(1/x)
tan(y)=(1/x)=O/A
O=1, A=x
{{{H=sqrt(O^2+x^2)=sqrt(1+x^2)}}}
{{{cos(y)=A/H=x/sqrt(1+x^2)}}}
{{{cos(2y)=cos^2(y)-sin^2(y)=x^2/(x^2+1)-1/(x^2+1)=(x^2-1)/(x^2+1)=(x+1)(x-1)/(x^2+1)}}}