# Prove trig identity: $\tan + \cot = \sec \csc$

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}

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