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

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

cosx / (1+sinx) ?

yeah that is LHS

multiply the numerator and denominator by cosx

oh yea

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

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

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

(1-sinx) times (1+sinx) = |dw:1325741008297:dw|