F.4 Maths 2!!!

1. tan(3π/2+A)tan(π+A)
=1/tanA*(-tanA)
=-1

2.sin(π/2+A)cos(2π-A)+cos(A-π/2)sin(π+A)
=cosAcosA+sinA(-sinA)
=cos^2 A-sin^2 A
=cos(2A)

did you type it wrong?

3.sinA(sinA)-cosA(cosA)
=sin^2 A-cos^2 A
=-cos2A

-1/tanA X tanA
= --------------------------------------
(cosA)^2-(sinA)^2
1
=__________________________
sinA(sinA)-cosA(cosA)

Please!!!Help
me!!!!!!!

the question:

Simplify this:

Tan(3丌/2+A)Tan(丌+A)

______________________________________

Sin(丌/2+A)Cos(2丌－A)+Cos(A－丌/2)Sin(丌+A)

the answer is 1

__________________________

SinA(SinA)－CosA(CosA)

what is the 過程?

THX!!!!!

Sol

………….Tan(3丌/2+A)Tan(丌+A)

-------------------------------------------------------------

Sin(丌/2+A)Cos(2丌－A)+Cos(A－丌/2)Sin(丌+A)



..… [－Cot(A)][Tan(A)]

=--------------------------------------------

[Cos(A)][Cos(A)]+[Sin(A)][－Sin(A)]



…………….1

=-------------------------------

SinA(SinA)－CosA(CosA)

2010-04-17 23:19:01 補充：
Tan(3丌/2+A)=－Cot(A)
Tan(丌+A)=Tan(A)
Sin(丌/2+A)=Cos(A)
Cos(2丌－A)=Cos(A)
Cos(A－丌/2)=Sin(A)
Sin(丌+A)=－Sin(A)