How Do You Solve These Trigonometric Identities?

[fluffertoes]

How do you do this one? I can't figure it out!
(2 - 5cot x) / (2 + 5cos x) = (2sin x - 5cos x) / (2sin x + 5cos x)

 

[MarkFL]

Are you sure you've copied it correctly? It appears to me that the actual identity should be:

\(\displaystyle \frac{2-5\cot(x)}{2+5\cot(x)}=\frac{2\sin(x)-5\cos(x)}{2\sin(x)+5\cos(x)}\)

 

[JorgeLuna]

How do you do this one? I can't figure it out!
(2 - 5cot x) / (2 + 5cot x) = (2sin x - 5cos x) / (2sin x + 5cos x)

fixedtwo ways to go

1) multiply left side by sinx/sinx

or

2) multiply right side by cscx/cscx

recall cotx = cosx/sinx and cscx = 1/sinx