math Question~~~~~~~

√(1+tan^2θ)=1/cosθ
[SInce sin^2 θ + cos^2 θ = 1]
[tan^2 θ + 1 = 1/cos^2 θ]
L.H.S.
= √(1+tan^2 θ)
= √(1/cos^2 θ)
= 1/cosθ
= R.H.S.

5+4sin(90°-θ)-sin^2θ=(cosθ+2)^2
L.H.S.
= 5+4sin(90°-θ)-sin^2θ
= 5 + 4cosθ - (1 - cos^2 θ)
= 5 + 4 cosθ - 1 + cos^2 θ
= cos^2 θ + 4 cosθ + 4
= (cosθ + 2)^2
= R.H.S.


2010-08-07 21:22:48 補充：
sin^2 θ + cos^2 θ = 1
tan^2 θ + 1 = 1/cos^2 θ <<<< 全部除 cos^2 θ
tan^2 θ + 1 = sec^2 θ

1+tan^2θ = sec^2θ

LHS:
√(1+tan^2θ)
sqrt(cos^2x+sin^2x)/(cos^2x))
sqrt1/sqrtcos^2x
1/cosx
_______________
RHS:

1/cosx

They are an identity
________________
5+4sin(90°-θ)-sin^2θ=(cosθ+2)^2
___________
LHS:
5+4sin(90°-θ)-sin^2θ
5+4cosx-sin^2x
______________
(cosθ+2)^2
cos^2x+4cosx+4
_________________

cos^2x+4cosx+4=5+4cosx-sin^2x
(cos^2x+sin^2x)=5-4
cos^2x+sin^2x=1

√(1+tan^2θ)<>1/cosθ