asin2x+bcos2x=m 1
bsin2x+acos2x=n 2
in 1 and 2 take cos2x common and send it to rhs in both equations then the resulting equations would be
atan2x+b=msec2x
=>atan2x+b=m+mtan2x
=>tan2x=m-b/a-m 3
similarly second equation would be
tan2y=n-a/b-n 4
given atanx=btany
squaring on both sides and regrouping we get
tan2x/tan2y=b2/a2 5
divide 3 and 4 and equte it to 5
we get
(m-b)(n-b)/(n-a)(m-a)=b2/a2
by cross multiplication and regrouping of the above equation we get
1/m+1/n=1/a+1/b
tan((x+y)/2)=sinx+siny/cosx+cosy
=a/b
sin(x+y)=2tan((x+y)/2)/1+tan2((x+y)/2)
substituting the abve value we get
sin(x+y)=2a/1+b2
