Proving trigonometric identities

I’ll use x instead of theta. Also keep in mind (sin(x))/(cos(x)) = tan(x)

i) sin(x) = o/h, cos(x) = a/h
Therefore sin^2(x) + cos^2(x) = (o^2 + a^2)/(h^2)
But because of Pythagoras o^2 + a^2 = h^2
So sin^2(x) + cos^2(x) = (h^2)/(h^2) = 1
Rearranging 1 - cos^2(x) = sin^2(x)

ii) Rearranging i), 1 - sin^2(x) = cos^2(x)
Now times both sides by tan(x).
Because tan(x)cos(x) = sin(x) one of the cos(x)s in cos^2(x) becomes a sin(x)
Therefore (1 - sin^2(x))tan(x) = cos(x)sin(x)

iii) Again i) says 1 - cos^2(x) = sin^2(x)
Now divide each term by (sin^2(x))
Done.

iv) Similar to iii) except divide everything by (cos^2(x)) instead of (sin^2(x)).

thanks very much! A lot simpler than I thought.

Okay can you tell me why/how:

1/cos^2(x) + 1/sin^2(x) = 1/cos^2(x)sin^2(x) ?