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Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$

I tried expanding $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)$ and expanding $tanfracX2tanfracY2$ and then cross multiplying , but I can't seem to find the answer.

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*