Hannah_Ahn
prove the identity:
1 over sec x + tan x
=
1  sin x over cos x



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Hannah_Ahn
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\[\left( 1 \over secx + tanx \right) = \left( 1sinx \over \cos x \right)\]

Yifan12879
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LHS=cosx / (1+sinx)
LHS/RHS = (cos^2 x)/(1sin^2 x)=cos^2 x / cos^2 x =1
so LHS=RHS

Hannah_Ahn
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cosx / (1+sinx) ?

Yifan12879
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yeah that is LHS

Yifan12879
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multiply the numerator and denominator by cosx

Hannah_Ahn
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oh yea

Hannah_Ahn
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I don't get LHS/RHS? what's going on
hehe

Yifan12879
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divide LHS by RHS and found it is 1.
so we can say LHS=RHS

Hannah_Ahn
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(1sinx) times (1+sinx) is one?

Yifan12879
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(1sinx) times (1+sinx) = dw:1325741008297:dw