$\displaystyle \frac{\sin[x]}{\tan[x]} +\cos[x]=\sec[x]$ I can get this to be 2 cos x = sec x I cannot see how to get those equal
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Originally Posted by IDontunderstand $\displaystyle \frac{\sin[x]}{\tan[x]} +\cos[x]=\sec[x]$ $\displaystyle \displaystyle\frac{\sin{x}}{\frac{\sin{x}}{\cos{x} }}+\cos{x}=\cos{x}+\cos{x}=2\cos{x}$ I don't know how you are going to get sex{x} out of that though
Originally Posted by dwsmith $\displaystyle \displaystyle\frac{\sin{x}}{\frac{\sin{x}}{\cos{x} }}+\cos{x}=\cos{x}+\cos{x}=2\cos{x}$ I don't know how you are going to get sex{x} out of that though Thank you. That is what I was thinking. Maybe it is a typo.
I agree with dwsmith. If you plug in $\displaystyle x=20^{\circ},$ you get an inequality.
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