Prove the identity: tan(x/2) + cot(x/2) = 2cscx?

Question

I've been trying forever...please help!

Answer 1

Use half-angle identity: tan(x/2) = (1 - cos x)/sin x = sin x/(1 + cos x).

Then cot(x/2) = 1/tan(x/2) = (1+cos x)/sin x, so

tan(x/2) + cot(x/2) = (1-cos x)/sin x + (1+cos x)/sin x
= (1-cos x+1+cos x)/sin x
= 2/sin x
= 2 csc x.

Answer 2

one can keep it simple
let x/2 = t

we have LHS = tan t + 1/tan t = (tan^2 t+1)/ tan t
= sec^ t/ tan t
= 1/cos ^2 t /( sint/ cos t)
= 1/(sin t cos t) = 2/ sin 2t
= 2 cosec 2t = 2 cosec x
= RHS

Answer 3

tan(x/2) + cot(x/2) =
(sin(x/2) / cos(x/2)) + (cos(x/2) / sin(x/2)) =
((sin(x/2))^2 / (sin(x/2)cos(x/2))) + ((cos(x/2))^2 / (cos(x/2)sin(x/2))) =
(sin^2(x/2)+cos^2(x/2)) / (cos(x/2)sin(x/2))=
1 / (cos(x/2)sin(x/2)) =
2 / (2cos(x/2)sin(x/2)) =
2 / sin(x) =
2csc(x)