椭圆与双曲线之间有许多类似的性质，
P是椭圆x^2/a^2+y^2/b^2=1(a>b>0)上任一点，焦点F1，F2，∠F1PF2=α，
三角形F1PF2面积为b^2sinα/(1+cosα)，类比，
P是双曲线x^2/a^2-y^2/b^2=1(a>0，b>0)上任一点，焦点F1，F2，∠F1PF2=α，
则三角形F1PF2面积为______
答案是b^2sinα/(1-cosα),求具体过程！
令PF1=X1>PF2=X2
Cosα=(X1^2+X2^2-F1F2^2)/2X1X2
2X1X2Cosα=(X1^2+X2^2-2X1X2)+2X1X2-4c^2
2X1X2Cosα=(X1-X2)^2+2X1X2-4c^2
2X1X2Cosα=4a^2+2X1X2-4c^2
X1X2(Cosα-1)=2a^2-2c^2=-2b^2
X1X2=2b^2/(1-Cosα)
S△F1PF2
=(X1X2*Sinα)/2
=2b^2*Sinα/2(1-Cosα)
=b^2sinα/(1-cosα)
Cosα=(X1^2+X2^2-F1F2^2)/2X1X2
2X1X2Cosα=(X1^2+X2^2-2X1X2)+2X1X2-4c^2
2X1X2Cosα=(X1-X2)^2+2X1X2-4c^2
2X1X2Cosα=4a^2+2X1X2-4c^2
X1X2(Cosα-1)=2a^2-2c^2=-2b^2
X1X2=2b^2/(1-Cosα)
S△F1PF2
=(X1X2*Sinα)/2
=2b^2*Sinα/2(1-Cosα)
=b^2sinα/(1-cosα)
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