# Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined

I appreciate the help.

My attempt:

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

• 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.) Commented Feb 3, 2014 at 0:48
• yup. It's quicker to go from $\frac1{cos\cdot{sin}}$ to $\frac1{cos}\frac1{sin}=sec\cdot{csc}$. Commented Feb 3, 2014 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 $$\begin{array}{lll} \tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\ &=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\ &=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\ &=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\ &=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\ &=&\sec\theta\csc\theta \end{array}$$

For acute $$\theta$$, there's this trigonograph:

$$\sec\theta \cdot \csc\theta \;=\; 2\,|\triangle OPQ| \;=\; 1\cdot( \tan\theta +\cot\theta)$$

(Showing that this works in any quadrant is straightforward.)

An alternative approach writes $$t=\tan x/2$$ so $$\tan x+\cot x=\frac{2t}{1-t^2}+\frac{1-t^2}{2t}=\frac{1+t^2}{1-t^2}\frac{1+t^2}{2t}=\sec x\csc x.$$

Equations last but one, last but two and last but three can be deleted and you can straight away jump to last result because you already know $$\sec,\csc$$ are inverse trig functions of $$\cos, \sin$$.