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anonymous
 one year ago
rewrite terms in sine and cosine only:
sec X  cosX/(1+sin X)
anonymous
 one year ago
rewrite terms in sine and cosine only: sec X  cosX/(1+sin X)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we can substitute this: \[\sec x = \frac{1}{{\cos x}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so we can write: \[\sec x  \frac{{\cos x}}{{1 + \sin x}} = \frac{1}{{\cos x}}  \frac{{\cos x}}{{1 + \sin x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would i simplify it?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3hint: the common denominator is: \[\cos x\left( {1 + \sin x} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3what is: \[\frac{{\cos x\left( {1 + \sin x} \right)}}{{\cos x}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how is cos the denominator?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is an intermediate step

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the denominator of our new equivalent fraction is: \[{\cos x\left( {1 + \sin x} \right)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then i get \[\frac{1+\sin X\cos ^{2}X\ }{ cosX(1+sinX) }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, we can use this identity: \[{\left( {\cos x} \right)^2} = \left( {1  \sin x} \right)\left( {1 + \sin x} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yay thank you so much !!!
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