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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 ^_^
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
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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 :)