$\displaystyle \frac{\sin[x]}{\tan[x]} +\cos[x]=\sec[x]$

I can get this to be 2 cos x = sec x

I cannot see how to get those equal

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- Jan 17th 2011, 05:43 PMIDontunderstandEstablish the following Identities
$\displaystyle \frac{\sin[x]}{\tan[x]} +\cos[x]=\sec[x]$

I can get this to be 2 cos x = sec x

I cannot see how to get those equal - Jan 17th 2011, 05:49 PMdwsmith
- Jan 17th 2011, 06:29 PMIDontunderstand
- Jan 18th 2011, 01:45 AMAckbeet
I agree with dwsmith. If you plug in $\displaystyle x=20^{\circ},$ you get an inequality.