Can you solve this trig identity

show that sinx/1-sinx - sinx/1 + sinx is equivalent to 2tan^2x
...

(sinx/1-sinx) - (sinx/1 + sinx) = 2tan^2x
LHS
(sinx/1-sinx) - (sinx/1 + sinx)

Get the LCM of the the two....(1-sinx) (1+sinx)
Then simply just as you would do with a regular fraction.
You would get
sinx + sin^2x - sinx + sin^2x

2sin^2x/(1-sinx) (1+sinx)
Now divide by (1+sinx)
Then you will remain with
2sin^2x/1-sinx
1-sinx is the same as cos^x
So 2sin^2x/cos^2x
Tan is sinx/cosx
So 2tan^2x Q.E.D