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OpenStudy (anonymous):
\[Let ~~ f(a)={cosA \over 1 +sinA}+{1+sinA \over cosA}\]\[(a) ~Prove ~that ~~f(A)=2secA\]
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OpenStudy (anonymous):
multiply the first fraction by (1-sinA) both num and denom one cos A gets cancelled then u may have (1+sinA+1-sinA)/cosA = 2secA
OpenStudy (anonymous):
Hm. Ok. that makes sense.
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