原式=
| sinAcosC+cosAsinC |
| sinBcosC+cosBsinC |
| sin(A+C) |
| sin(B+C) |
| sinB |
| sinA |
| b |
| a |
∵aq+aq2>a,①
a+aq>aq2②
a+aq2>aq,③
解三个不等式可得q >
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| 2 |
0 <q<
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| 2 |
综上有
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| 2 |
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| 2 |
故答案为(
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| 2 |
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| 2 |
| sinA+cosAtanC |
| sinB+cosBtanC |
| sinAcosC+cosAsinC |
| sinBcosC+cosBsinC |
| sin(A+C) |
| sin(B+C) |
| sinB |
| sinA |
| b |
| a |
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| 2 |
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| 2 |
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| 2 |
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| 2 |
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| 2 |
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| 2 |