Question

Prove both sides equal: (Sin x/1-cosx)+(sin x/1+cos x)=2 csc x

Answer 1

Woops, it shouldn't say simplify.
It should say "prove this" or something. Bah whatever, same thing :)
So let's start with this side.
\[\large \frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}\]

We'll start by getting a common denominator.
To do so, we'll simply multiply one base by the other.\[\large \color{royalblue}{\left(\frac{1+\cos x}{1+\cos x}\right)}\frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}\color{royalblue}{\left(\frac{1-\cos x}{1-\cos x}\right)}\]

Answer 2

Fixed that, lol

Answer 3

Simplifying the denominators gives us,\[\large \frac{\sin x(1+\cos x)}{1-\cos^2 x}+\frac{\sin x(1-\cos x)}{1-\cos^2 x}\]

Answer 4

Since they have a common denominator, we can combine the fractions,\[\large \frac{\sin x(1+\cos x)+\sin x(1-\cos x)}{1-\cos^2 x}\]

Answer 5

Factoring out a sine from each term in the top gives us,
\[\large \frac{\sin x\left(1+\cos x+1-\cos x\right)}{1-\cos^2 x}\]

Answer 6

\[\large \frac{\sin x\left(1\cancel{+\cos x}+1\cancel{-\cos x}\right)}{1-\cos^2 x} \quad = \quad \frac{2\sin x}{1-\cos^2x}\]

Answer 7

From here, we'll want to remember our most basic trig identity.\[\large \color{salmon}{\sin^2x+\cos^2x=1} \qquad \rightarrow \qquad \color{salmon}{1-\cos^2x=\sin^2x}\]

Answer 8

\[\large \frac{2\sin x}{\color{salmon}{1-\cos^2x}} \qquad \rightarrow \qquad \large \frac{2\sin x}{\color{salmon}{\sin^2x}}\]

Answer 9

\[\large \frac{2\cancel{\sin x}}{\color{salmon}{\sin^{\color{}{\cancel2}1}x}} \qquad = \qquad \frac{2}{\sin x} \qquad = \qquad 2\csc x\]

Answer 10

It's a A LOT of little steps.
Let me know if you got confused on any of them.

Answer 11

I'm good.

Answer 12

THANKS!