# Proving the identity $\frac{1}{\tan x}+\tan x=\frac1{\sin x\cos x}$

As I'm working on my trigonometry exercises, I come across an exercise called "prove the equality," which I proceed to solve. This all goes well, until I arrive at a certain point, after which I can't solve any further and don't understand the explanation offered in the answers section. If you would please help me understand the way of thinking needed to solve this question, I'd be very happy!

Here's the problem:

Solve the equality:

$$\frac1{\tan(x)} + \tan(x) = \frac1{\sin(x)\cdot\cos(x)}$$

And here's what I've solved for:

$$\text{RHS}=\frac{1}{\frac{\sin(x)}{\cos(x)}} + \frac{\sin(x)}{\cos(x)} =\frac{\cos x}{\sin x} +\frac{\sin(x)}{\cos(x)}$$

Then, I get stuck. The answers section says the next step should be:

$$\frac{\cos^2(x) + \sin^2(x)}{\sin(x) \cdot \cos(x)} = \text{RHS}$$

I understand how this proves the equality, but I don't understand how they just went from the previous step to this one: where do the exponents suddenly come from??

If you would be so kind as to explain this sudden raise of powers to me... My gratitude will be infinite!

Lila

• Welcome to Math.SE! Please use MathJax. For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Commented May 3, 2018 at 10:23
• Thanks! I'll definitely do so in the future
– Lila
Commented May 3, 2018 at 10:26
• $$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$$
– user284001
Commented May 3, 2018 at 10:28
• @Kevin Thanks! Will bookmark this one for sure ;)
– Lila
Commented May 3, 2018 at 10:30
• @Lila No problem, I presume your equality only holds for $x \neq \pi/2$ (plus the periodic zeroes thereafter). In the identity above, you cannot ever have $b, d =0$.
– user284001
Commented May 3, 2018 at 10:31

Way of thinking: You want to simplify the more complicated side, which is the LHS, to a less complicated expression. You have two fractions with different denominators. So you multiply the numerator and denominators of the two fractions to obtain a common denominator.$$\frac1{\tan x}+\tan x=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\frac{\cos^2 x}{\cos x\sin x}+\frac{\sin^2 x}{\sin x\cos x}=\frac{\cos^2 +\sin^2x}{\sin x\cos x}$$
• No problem! ${}$ Commented May 3, 2018 at 10:30
$$\frac1{\tan x}+\tan x=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\frac{\cos^2 x+\sin^2 x}{\sin x\cos x}$$