In triangle ABC,
show that cosAcosecB + secAsinB = [1+cos^2 A -cos^2(A+C)]/cosAsin(A+C)
simple maths
2013-06-10 4:23 am
回答 (3)
2013-06-11 12:33 am
✔ 最佳答案
LHS= cosAcosecB + secAsinB
= cosA/sinB + sinB/cosA
= (cos^2 A +sin^2 B)/ cosAsinB
= (cos^2 A +1-cos^2 B)/ cos A sin[π-(A+C)]
= {1+cos^2 A- cos^2 [π-(A+C)]}/ cosAsin[π-(A+C)]
= 1+cos^2 A- cos^2(A+C)/ cosAsin(A+C)
= RHS
Hope I can help you!
2013-06-10 6:02 am
[1 + cos^2 A - cos^2 (A+ C)]/[cos A sin (A + C)]
= (1 + cos^2 A - cos^2 B)/(cos A sin B)
= (sin^2 B + cos^2 A)/(cos A sin B)
= sin B/cos A + cos A/sin B
= cos A csc B + sin B sec A
= (1 + cos^2 A - cos^2 B)/(cos A sin B)
= (sin^2 B + cos^2 A)/(cos A sin B)
= sin B/cos A + cos A/sin B
= cos A csc B + sin B sec A
參考: Myself
2013-06-10 4:41 am
Hint: Start from RHS, change "A+C things" to "B things".
收錄日期: 2021-04-13 19:30:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130609000051KK00266