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

That question makes absolutely no sense.. could someone help me? or lead me in the direction to figuring it out?

and..
(Beside the trig functions is theta)

1+1/cos = tan^2/sec-1

maybe you mean

sin t/(1-cos t) + (1-cos t)/sin t

sin^2 t (1-cos t) (1-cos t)^2
----------------- + ----------------
(1-cos t)(sin t) ______ (1-cos t)(sin t)

numerator only for a while

sin^2 t-sin^2 t cos t +1 -2 cos t+cos^2 t
but sin^2 + cos^2 = 1
2 -sin^2 t cos t - 2 cos t
put denominator back
2 -sin^2 t cos t - 2 cos t
--------------------------
(1-cos t)(sin t)

I think I misunderstood your lack of parentheses but I think you can get the idea
BE CAREFUL ABOUT PARENTHESES!!
when typing on the computer i can not tell what you mean for numerators or denominators.

sin t/(1-cos t) + (1-cos t)/sin t

sin^2 t ______________ (1-cos t)^2
----------------- + ----------------
(1-cos t)(sin t) ______ (1-cos t)(sin t)

sin^2 t + 1 -2 cos t +cos^2 t
--------------------------------
(1-cos t)(sin t)

2 - 2 cos t
--------------
(1-cos t) sin t

= 2/sin t
which is csc t

The parenthesis were for the arguments.. My question was how do I verify this indentity

Theta = argument..

I did verify it.

I only did the left side, starting with what you gave me and ending with
2/sin t
which is 2 csc t
which is the right side

Here is how you should have written it

(use t for theta)

sin(theta)/ { 1-cos(theta)} + {1-cos(theta)}/sin(theta) = 2 csc(theta)

The critical point is that afteter you get it all over the LCD (sin t)(1-cos t)

you see that you have sin^2 t+ cos^2 t which is one in the numerator.

I don't understand what you did?

How would x'ing the right side of the plus side by sin give you (1-cos)^2

OH nevermind about that .