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

I appreciate the help.

My attempt:

$$ \begin{align} \tan + \cot &= \frac{\sin}{\cos} + \frac{\cos}{\sin} \\ &= \frac{\sin^2}{\cos \sin}+\frac{\cos^2}{\cos \sin} \\ &= \frac{\sin^2+\cos^2}{\cos \sin}\\ &= \frac{1}{\cos \sin}\\ &= \frac{1}{\frac{1}{\sec}\frac{1}{\csc}}\\ &=\frac{1}{\frac{1}{\sec \csc}}\\ &=\frac{1}{1}\cdot \frac{\sec \csc}{1}\\ &= \sec \csc \end{align} $$

share|cite|improve this question
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ \ \theta \ $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.) – RecklessReckoner Feb 3 '14 at 0:48
yup. It's quicker to go from $\frac1{cos\cdot{sin}}$ to $\frac1{cos}\frac1{sin}=sec\cdot{csc}$. – Eleven-Eleven Feb 3 '14 at 0:49

That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.

Finally, you could save time on your proof by noticing on the fourth step that $$ \frac{1}{\cos x\sin x}=\frac{1}{\cos x}\frac{1}{\sin x}=\sec x \csc x $$

share|cite|improve this answer

Your steps are correct, but just keep in mind that robotically converting everying into $\sin$s and $\cos$s isn't the only option available to you.

Note that $$\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}=\frac{\csc\theta}{\sec\theta}$$ that $$\cot\theta\tan\theta=\frac{1}{\tan\theta}\cdot\tan\theta=1$$ and that $$\sec^2\theta=\tan^2+1$$ then $$\begin{array}{lll} \tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\ &=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\ &=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\ &=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\ &=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\ &=&\sec\theta\csc\theta \end{array}$$

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.