Question

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 1

\[((sinx)/(1-cosx)) + ((sinx)/(1+cosx)) = 2 \csc x\]

Answer 2

Ooo these are fun.
Hmmm let's see..
You remember how to do stuff with cojugates? (a-b)(a+b)

Answer 3

If we multiply the first fraction, top and bottom, by the `conjugate` of the denominator,
and then do the same with the other fraction, we can get a nice common denominator.

Answer 4

\(\large\rm \dfrac{\sin x}{1-\cos x}\color{royalblue}{\left(\dfrac{1+\cos x}{1+\cos x}\right)}+\dfrac{\sin x}{1+\cos x}\color{orangered}{\left(\dfrac{1-\cos x}{1-\cos x}\right)}\)

Answer 5

Don't multiply out the tops, leave the brackets.
Multiplying out the bottoms gives us the difference of squares \(\Large\rm 1-\cos^2x\).

Do you remember your Pythagorean Identity for Sine and Cosine?

Answer 6

sin=o/ad
cos=a/h
@zepdrix

Answer 7

No no no you silly billy :o
Our Pythagorean Identity is:
\(\Large\rm \sin^2x+\cos^2x=1\)
If we subtract cos^2x from each side it gives us,
\(\Large\rm \sin^2x=1-\cos^2x\)
Do you see how we can maybe use this? :o

Answer 8

oh yea oops wrong one :/

Answer 9

After multiplying out the conjugates our equation is currently,
\(\Large\rm \dfrac{\sin x(1+\cos x)}{1-\cos^2x}+\dfrac{\sin x(1-\cos x)}{1-\cos^2x}\)

Answer 10

Would I somehow combine them?

Answer 11

Not just yet.
The Pythagorean Identity allows us to convert our denominators to sin^2x, yes?

Answer 12

yes since sin^2x=1−cos^2x

Answer 13

After that you can cancel out a sin from each top and bottom, leaving you with:
\(\Large\rm \dfrac{1+\cos x}{\sin x}+\dfrac{1-\cos x}{\sin x}\)And from THIS place, it would be a good idea to combine.

Answer 14

\(\Large\rm \dfrac{1+\cos x+1-\cos x}{\sin x}\)

Answer 15

What should we do next? :o
What do you think?

Answer 16

2/sinx?

Answer 17

Ok good.
Let's pull the 2 off of the fraction so maybe you'll remember another identity we can use.

\(\Large\rm 2\left(\dfrac{1}{\sin x}\right)\)

Answer 18

Bah I've gotta get going :c