Question

(sin x / 1 - cos x ) + ( sin x / 1+ cos x ) = 2 csc x

Answer 1

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 2

How far did you get?

Answer 3

done even know where to start . @e.mccormick

Answer 4

No need to @ tag a person when they are right here. Hehe.

Well, you are already in sine and cosine on the left, so what could you do to add the fractions?

Answer 5

i need to get the same denominators .

Answer 6

Yep!

Answer 7

and how do i do that

Answer 8

I assume you meant this:
\(\dfrac{\sin x}{1 - \cos x} + \dfrac{ \sin x}{ 1 + \cos x} = 2 \csc x\)

Answer 9

yes , so how do i make the long part look like the short side ?

Answer 10

Well, just because it is a trig function does not make the fraction at all special. You use the same way of finding a common denominator as you do with any fraction. Multiply by the missing part over the missing part!

Answer 11

i know what i need to do , i just dont know how

Answer 12

The timplest case is with just numbers:
\(\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{2}\cdot\dfrac{3}{3}+\dfrac{1}{3}\cdot\dfrac{2}{2}=\dfrac{3}{6}+\dfrac{2}{6}\)

It is the same process as that. Because 1/2 does not have the 3 in it, I multiply by 3/3. Etc.

Answer 13

I know how to add fractions.

Answer 14

Having variables or other things in there also does not change this process:
\(\dfrac{1}{2+x}+\dfrac{1}{3+x}=\)

\(\dfrac{1}{2+x}\cdot\dfrac{3+x}{3+x}+\dfrac{1}{3+x}\cdot\dfrac{2+x}{2+x}\)

It is exactly the same for trig fractions. So if you know how to add fractions, you know how to add these.

Answer 15

@genius12 help please ?