prove that tg(x) + ctg(x) = 2 cosec(2x)

First, we know that

cosec x = 1/sin x, so that cosec (2x) = 1/sin2x

We'll re-write the equation:

tg(x) + ctg(x) = 2/sin2x

We'll divide by 2 both sides:

[tg(x) + ctg(x)]/2 = 1/sin2x

But ctg x = 1/tg x

So, tg(x) + ctg(x)  = tg x + 1/tg x

We'll find the common denominator and we'll add:

tg(x) + ctg(x) = [(tg x)^2 + 1]/tgx

But (tg x)^2 + 1 = 1/(cosx)^2

[(tg x)^2 + 1]/tgx = 1/(tg x)*(cosx)^2

1/(tg x)*(cosx)^2 = cos x/(sin x)*(cos x)^2

After reducing similar terms, we'll get:

cos x/(sin x)*(cos x)^2 = 1/sin x * cos x

[tg(x) + ctg(x)]/2 = 1/2*sin x * cos x = 1/sin 2x

But, from enunciation, [tg(x) + ctg(x)]/2 = 1/sin2x, so the result is verified.