anonymous
  • anonymous
prove the identity: 1 over sec x + tan x = 1 - sin x over cos x
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\left( 1 \over secx + tanx \right) = \left( 1-sinx \over \cos x \right)\]
anonymous
  • anonymous
LHS=cosx / (1+sinx) LHS/RHS = (cos^2 x)/(1-sin^2 x)=cos^2 x / cos^2 x =1 so LHS=RHS
anonymous
  • anonymous
cosx / (1+sinx) ?

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anonymous
  • anonymous
yeah that is LHS
anonymous
  • anonymous
multiply the numerator and denominator by cosx
anonymous
  • anonymous
oh yea
anonymous
  • anonymous
I don't get LHS/RHS? what's going on hehe
anonymous
  • anonymous
divide LHS by RHS and found it is 1. so we can say LHS=RHS
anonymous
  • anonymous
(1-sinx) times (1+sinx) is one?
anonymous
  • anonymous
(1-sinx) times (1+sinx) = |dw:1325741008297:dw|

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