[MrRubix]

I'm a little rusty when it comes to trig identities but here goes:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

First I'd combine the first two terms:

(cos^2X+sin^2X)/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

The numerator, cosX^2 + sin^2X, is equal to 1:

1/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

This is the same as:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

So now we are basically trying to equate the things within the second-power-raise, so we aim to equate 1/(sinX*cosX) and tan X + 1/tanX.

Since tan = sin/cos and 1/tan or cot = cos/sin, we see that the right hand side is basically sin/cos + cos/sin, so we have two instances of each. Currently, the left hand side has half as many instances. Therefore we should try to prove this by simplifying the right-hand side:

tanX + 1/tanX
We then change tanX into tan^2X/tanX so we can combine like terms:
(1 + tan^2X)/tanX
And change our tangents into their sine and cosine equivalents:
(1 + (sinX/cosX)^2)/(sinX/cosX)
We then change the 1 into cos^2X/cos^2X so we can combine it as a like term later:
((cos^2X/cos^2X) + (sin^2X/cos^2X))/(sinX/cosX)
Combining terms:
(((cos^2X + sin^2X)/cos^2X))/(sinX/cosX)
cos^2X + sin^2X is equal to 1:
(1/cos^2X)/(sinX/cosX)
Then we simplify things a bit:
cosX/(sinX*cos^2X)
And finally drop the common cosX from the numerator and denominator:
1/(sinX*cosX)

Since we just proved 1/(sinX*cosX) = tanX + 1/tanX, we therefore prove:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

And since we showed that (1/sin^2X) + (1/cos^2X) = (1/(sinX*cosX))^2, we therefore prove:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

Edit: Oh yeah, QED