take the reciprocal of 3/ cos^2 theta?
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Yes, so we now have: $\displaystyle \sqrt{\frac{1}{\cos^2(\theta)}+1}\cdot\frac{\cos^2 (\theta)}{3}=\frac{\cos^2(\theta)\sqrt{\frac{1}{ \cos^2(\theta)}+1}}{3}$
now what?
You could take the square of the cosine function under the radical and distribute...
Originally Posted by mathisfun26 now what? ... ... you have a mess. whoever came up with this problem should be slapped, or forced to find the decimal representation of $\displaystyle \sqrt{2}$ to one thousand decimal places ... by hand.
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