Question

Prove it is true: Trigonometric Solving

Answer 1

\[Sinx/1-\cos x+\sin x/1+\cos x=2\csc x\]

Answer 2

I don't know if anyone can see that but here is the raw equation (I can't see it in equation form). Sin x/1-cos x + sin x/1 + cos x = 2csc x

Answer 3

\[\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx }=2cscx\]

Answer 4

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

(I'm repeating what I can see from that so you can tell what I see from it)

Answer 5

Okay, so first what you do is get a common denominator: 1-cos^2x:
\[\frac{ sinx(1+cosx) }{ 1-cos^2x}+\frac{ sinx(1-cosx) }{ 1-\cos^2x }\]
Now distribute the top and combine like terms:
\[\frac{ sinx+sinxcosx+sinx-sinxcosx }{ 1-\cos^2x }\]
The sinxcosx cancel one another and you are left with:\[\frac{ 2sinx }{ 1-\cos^2x }\]
Now using trig identities we can substitute sin^2x for 1-cos^2x:
\[\frac{ 2sinx }{ \sin^2x }\]
Now we can cancel out a sinx from both top and bottom:
\[\frac{ 2 }{ sinx }\]
And guess what that equals?
\[2*\frac{ 1 }{ sinx }=2cscx\]

Answer 6

Does it make sense?

Answer 7

So far it is, I just have to re-write it so that I can understand it (when you use the "Equation" button it doesn't transfer over for me. But here's your medal, I'm almost certain you are correct. Thank you =)

Answer 8

Any time :P