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write sinx in termss of secx
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clue secx=(1/cosx)
Another hint: sin²x + cos²x = 1
\(\sec x = \dfrac{1}{\cos x}\) so \(\sec^2 x = \dfrac{1}{\cos^2 x}\)
Still don't get it? @Leemah!
Not really, I have just started my trig and precal class. I had straight A's in all my precious math classes, but I am really having difficulty with identities and trig in general. I have never seen it before. I have the answer as \[\sqrt{\sec x-1}/\sec x\] but don't know how to get there.
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