How do I solve this proof?

Question

I'm completely stuck, I've tried a lot of things, nothing works. We're supposed to use the Pythagorean identities but those lead to nothing! Some help, please?

sin2θ + sin2θ cot2θ = 1

Answer 1

If you mean sin² θ + sin² θ cot² θ = 1, here's the proof...

sin² θ + sin² θ cot² θ = sin² θ (1 + cot² θ) = sin² θ (csc² θ) = sin² θ (1/(sin² θ)) = 1

Answer 2

It would help loads to realize that your notation is LOUSY. What you want is: sin^2 O (1+cot^2 O)=1. One of the identities is that 1+cot^2 O = csc^2 O. However csc^2 O = 1/(sin^2 O), so you wind up with sin^2 O / sin^2 O = 1.

Answer 3

sin2θ + sin2θ(cos2θ/sin2θ) = 1 cot2θ = (cos2θ/sin2θ)
sin2θ + cos2θ = 1 <--- this is a pythagorean identity

Answer 4

I think you mean

sin ^2 t + sin ^2 t cot^2 t = 1

LHS = sin^2 t + sin ^2 t . cos^2 t/sin ^2 t
= sin ^2 t + cos^2 t = 1
= RHS

Answer 5

since cot2x is the same as cos2x/sin2x, the above equation simplifies to sin2x+cos2x=1. Now square both sides:
sin^2(2x)+cos^2(2x)+2cos(2x)*sin(2x)=1
but this implies that 2cos(2x)sin(2x)=0--which is definitely not true. Therefore, what you are trying to prove is actually not true.

Answer 6

It's actually pretty simple.
I'll give you a hint.
a2+b2=c2
and
tan45 =1