開方(1+tan^2X)=
A. a/b
B.b/a
C.-b/a
D.+-b/a
E.+-a/b
更新1:
Please anwser and explain
f4 maths
回答 (2)
2009-05-18 9:32 am
✔ 最佳答案
(sinx)^2=1-(cosx)^2(sinx)^2=1-(a/b)^2 and (cosx)^2=(a/b)^2
√(1+(tanx)^2)
=√(1+[(sinx)^2/(cosx)^2])
=√(1+[(1-(a/b)^2)/(a/b)^2])
=√(1+(b^2-a^2)/a^2)
=√(b^2/a^2)
=b/a
Ans: B
2009-05-18 01:37:12 補充：
最後個2步應該係:
=√(b^2/a^2)
=│b/a│
=-b/a
Ans: C
2009-05-18 9:36 am
√(1 + tan2 X)
= √sec2 X
= -b/a (must be positive)
Ans : C
= √sec2 X
= -b/a (must be positive)
Ans : C
收錄日期: 2021-04-22 00:42:21
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