Question 977115
LHS= 
{{{tan(t)=sin(t)/cos(t)}}}
={{{2*sin(t)cos(t)/(2*cos^2(t))}}} [Multiplying Num. and Den. by 2*cos(t)]
={{{sin(2t)/(2*cos^2(t)-1+1)}}}
={{{sin(2t)/(cos(2t)+1)}}}
=RHS
Hence,
tan15 = sin30/(1+cos30)
={{{(1/2)/(1+sqrt(3)/2)}}}
={{{1/(2+sqrt(3))}}}
Also,
tan45 = sin90/(1+cos90)
={{{1/(1+0)}}}
= 1