Please help me with an easy trig identity

cos(2x) / sin(x) + sin(2x) / cos(x) = csc(x)

Prove that it is an identity.

I've been stuck on this for hours and would really appreciate it if anyone could help.

[(cos 2x)/sin x] + [(sin 2x)/cos x] = csc x

cos 2x = 1 - 2 sin^2 x

sin 2x = 2 sin x cos x

first term...[(1 - 2 sin^2 x)/sin x]

second term...[(2 sin x cos x)/cos x] = 2 sin x

LCD: sin x

so,

[(1 - 2 sin^2 x) + (2 sin^2 x)]/sin x = 1/sin x = csc x

QED

...if you have been stuck for "hours," then go to your teacher for help, immediately...this was not hard ! Study double angle identities...

 [cos(2x) ⁄ sin(x)] + [sin(2x) ⁄ cos(x)] = 1 ⁄ sin(x)

   ... common denominator ...

 [ cos(x) • cos(2x) + sin(x) • sin(2x) ]  ⁄  [sin(x) • cos(x)] = 1 ⁄ sin(x)

   ... cancel sin(x) ...

 [ cos(x) • cos(2x) + sin(x) • sin(2x) ]  ⁄  cos(x) = 1

   ... substitute double angle identities ...

 [ cos(x) • { cos²(x) − sin²(x) } + sin(x) • { 2sin(x) • cos(x) } ]  ⁄  cos(x) = 1

   ... cancel cos(x) ...

  cos²(x) − sin²(x)  + 2sin(x) • sin(x) = 1

    cos²(x) − sin²(x)  + 2sin²(x) = 1

          cos²(x) + sin²(x) = 1