- #1

karush

Gold Member

MHB

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And $p$ and $q$ are $90^o<\theta<180^o$ and $p>q$

Show that $\tan\left({\theta}\right)=\frac{q-p}{2\sqrt{qp}}$

I tried using $q=\frac{2\pi}{3 }$ and $p=\frac{5\pi}{6}$

But not...

To do this but theory I'm clueless