Jawabanmu

2014-05-03T10:37:49+07:00
L.H.S. = (sin A + 1-Cos A)/ (Sin A -1+ Cos A)
Multiply and divide by SinA + 1 + CosA
= [(sin A + 1-Cos A) *(SinA + 1 + CosA) ]/ [(Sin A -1+ Cos A) *(SinA + 1 + CosA) ]
= Applying the rule (x+y)(x-y) = x^2 - y^2
=> [(sinA+1)^2 - cos^2A]/ [(SinA + cosA)^2 - 1]
= [sin^2A + 2sinA + 1 - cos^2A]/[sin^2A + cos^2A + 2sinAcosA -1]
Applying the identity sin^2A+cos^2A= 1, 1-cos^2A = sin^2A and 1-sin^2A = cos^2A
= [sin^2A + 2sinA + sin^2A]/[1 + 2sinA cosA -1]
= (2sin^2A + 2sinA)/2sinAcosA
= 2sinA(1+sinA)/2sinAcosA
= (1+sinA)/cosA
= 1/cosA + sinA/cosA
= secA + tanA
=R.H.S.

semoga membantu ^_^
2014-05-03T10:51:05+07:00
Misal :
Buktikan 1- cos 2a + sin 2a per 1 + cos 2a + sin 2a = tan a?

Jwb :
1. diubah menjadi bentuk 

sin" a + cos" a - (cos" a - sin" a) + 2 sin a cos a 
______________________________________... 

sin" a + cos" a + (cos" a - sin" a) + 2 sin a cos a 



2 sin" a + 2 sin a cos a 
____________________ 

2 cos" a + 2 sin a cos a 



2 sin a ( sin a + cos a ) 
___________________ 

2 cos a ( cos a + sin a ) 

sin a 
____ 
cos a 

= tan a <--

aku kasih contoh digambar juga :)