kodam abhigna
Last Activity: 6 Years ago
as we all know (a^2-b^2)=(a+b)(a-b)----------------(1)
and tan(a+b)=(tana+tanb)/(1-tana * tanb);------------------(2)
tan(a-b)=(tan a-tanb)/(1+tana * tanb);------------------(3)
LHS:
=[tan^2(2a)-tan^2(a)]/[1-tan^2(2a)*tan^2(a)]
as per equation (1);
=[{tan(2a)+tan(a)}{tan(a)-tan(a)}] / [{1-tan(a)tan(2a)}{1+tan(a)tan(2a)}]
on rewriting ;
=[(tan2a+tana)/(1-tan2a * tana)] * [(tan2a-tana)/(1+tan2a * tana)]
rewriting the first term using equation(1) and second term using equation(3);
=tan(2a+a) * tan(2a-a)
=tan3a * tana
=tan3a tana
=RHS
therefore LHS=RHS;
hence proved.