已知sina/sinb=p cosa/cosb=q, 且p不等于正负1，q不等于0,求tga*tgb的值

已知sina/sinb=p cosa/cosb=q, 且p不等于正负1，q不等于0,求tga*tgb的值 怎么做？并且告诉我为什么会想到这么做

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∵sinA/sinB=p；cosA/cosB=q∴sinA=p*sinB；cosA=q*cosB∴sinA^2+cosA^2=(p*sinB)^2+(q*cosB)^2即1=p^2-(p^2)cosB^2+(q^2)*cosB^2解得：cosB^2=(1-p^2)/(p^2-q^2)；sinB^2=(2p^2-q^2-1)/(p^2-q^2)tgA*tgB=(sinA/cosA)*(sinB/cosB)=sinA*sinB/cosA*cosB=p*sinB^2/(q*cosB^2)=(p/q)(sinB^2/cosB^2)=(p/q)[(2p^2-q^2-1)/(1-p^2)]=p(2p^2-q^2-1)/q(1-p^2)