题目内容
求sec50°+tan10°的值．分析：把原式先转化成弦，再利用和差化积公式进行化简，最后可得出答案．
解答：解：sec50°+tan10°
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=2cos30°
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| 1 |
| cos50° |
| sin10° |
| cos10° |
=
| 2sin50° |
| 2sin50°cos50° |
| sin10° |
| cos10° |
=
| 2sin50° |
| sin100° |
| sin10° |
| cos10° |
=
| 2sin50° |
| cos10° |
| sin10° |
| cos10° |
=
| 2sin50°+sin10° |
| cos10° |
=
| sin50°+(sin50°+sin10°) |
| cos10° |
=
| sin50°+cos20° |
| cos10° |
=
| cos40°+cos20° |
| cos10° |
=
| 2cos30°cos10° |
| cos10° |
=2cos30°
=
| 3 |
点评：本题主要考查弦切互化的问题．要熟练掌握三角函数中的如和差化积、积化和差、倍角公式等常用公式．
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