# Trigonometric Identities To Prove

1. $\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$

Left Side:
\begin{align*} \tan\theta+\cot\theta={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={\sin^2\theta+\cos^2\theta\over\cos\theta\sin\theta} = \dfrac{1}{1\sin\theta\cos\theta} \end{align*} Right Side:
\begin{align*} \dfrac{2}{\sin2\theta}=\dfrac{2}{2\sin\theta\cos\theta}=\dfrac{1}{1\cos\theta\sin\theta} \end{align*}

I got it now. Thanks!

• Your first line under "left side" has a pretty substantial error in adding fractions. You need to find a common denominator. – The Chaz 2.0 Jul 15 '12 at 1:50
• That's not how you add fractions! Is $\frac{1}{3}+\frac{1}{5}$ equal to $\frac{1+1}{3+5}$? – Arturo Magidin Jul 15 '12 at 1:50
• @TheChaz Didn't notice that. Thank you. – Austin Broussard Jul 15 '12 at 1:51
• @ArturoMagidin I told you my brain is working too fast! I've been doing this stuff for hours! – Austin Broussard Jul 15 '12 at 1:54
• @Austin: Sorry, but I don't see how that error comes from "brain working to fast". If anything, it comes from "brain working too slow"... – Arturo Magidin Jul 15 '12 at 1:56

$$\tan(\theta) + \cot(\theta) = {\sin(\theta)\over \cos(\theta)} + {\cos(\theta)\over \sin(\theta)} = {\sin^2(\theta) + \cos^2(\theta) \over\cos(\theta)\sin(\theta)} = {1\over\sin(\theta)\cos(\theta)}.$$ Now avail yourself of the fact that $$\sin(2\theta) = 2\cos( \theta)\sin(\theta).$$
I'll just put together what you wrote... \begin{align*} \tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta \cdot\sin\theta} = \dfrac{1}{\cos \theta \cdot \sin \theta} \end{align*}