Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  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!

share|cite|improve this question
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
up vote 1 down vote accepted

$$\tan(\theta) + \cot(\theta) = {\sin(\theta)\over \cos(\theta)} + {\cos(\theta)\over \sin(\theta)} = {\sin^2(\theta) + \cos^2(\theta) \over\cos(\theta)\sin(\theta)} = {1\over\sin(\theta)\cos(\theta)}.$$ Now avail yourself of the fact that $$\sin(2\theta) = 2\cos( \theta)\sin(\theta).$$

share|cite|improve this answer

I'll just put together what you wrote... $$\begin{align*} \tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta \cdot\sin\theta} = \dfrac{1}{\cos \theta \cdot \sin \theta} \end{align*}$$

Where the penultimate inequality is what you should have written.

Can you take it from here?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.