Question 149422
Re-writing,
{{{(tanA-cotA)/(tanA+cotA)}}}
By quotient identities:
={{{((sinA/cosA)-(cosA/sinA))/((sinA/cosA)+(cosA/sinA))}}}
={{{((sin^2(A)-cos^2(A))/cross((cosA)(sinA)))/((sin^2(A)+cos^2(A))/cross((cosA)(sinA)))}}}
={{{(sin^2(A)-cos^2(A))/(sin^2(A)+cos^2(A))}}} = {{{(sin^2(A)-cos^2(A))/1}}}-->A
We have to remember the identity for "double angle formulas":
{{{cos2(mu)=cos^2(mu)-sin^2(mu)}}} --------------> B
*note" {{{(mu)=A}}}
So,
= {{{-cos2A}}},  FINAL ANSWER ----> see the negative (-) sign, bec. the final answer {{{A}}} is opposite the identity {{{B}}}.
Thank you,
Jojo