Proving equations (trig)

Again I just need some help with the last few questions on a maths assignment in high school. I simply need to prove some equations using trig identities. Here are the ones I have had trouble with:

tan^2A - sin^2A = tan^2A * sin^2A

sinA/cosA * (1 - cot^2A) + cosA/sinA * (1 - tan^2A) = 0

sinA * cosA * (tanA + cotA) = 1

1 - cos^2A/1 + sinA = sinA

Any help with these would be much appreciated as the assignement is now due tommorow. Thanks.

tan² - sin² = sin²/cos² - (sin²cos²)/cos² = sin²(1 - cos²)/cos² = sin²(sin²)/cos² = tan²×sin² QED

sin/cos × (1 - cot²) + cos/sin × (1 - tan²) =
sin/cos × (1 - cos²/sin²) + cos/sin × (1 - sin²/cos²) =
sin/cos × (sin² - cos²)/sin² + cos/sin × (cos² - sin²)/cos² =
(sin² - cos²)/(sin)(cos) + (cos² - sin²)/(cos)(sin) =
(sin² - cos² + cos² - sin²)/(cos)(sin) = 0 QED

sin × cos × (tan + cot) =
sin× cos × (sin/cos + cos/sin) =
sin²cos/cos + cos²sin/sin =
sin² + cos² = 1 QED

Is that supposed to be (1 - cos²)/(1 + sin) = sin ? Or 1 - cos²/(1 + sin) = sin ?
If it's the second,
1 - cos²/(1 + sin) =
1 - (1 - sin²)/(1 + sin) =
1 - (1 - sin)(1 + sin)/(1 + sin) =
1 - (1 - sin) = sin QED
{If it's the first, you're on your own.}

tan^2A - sin^2A = tan^2A * sin^2A, know that tan^2A = sin^2A / cos^2A
Divide both sides by sin^2A: 1/cos^2A - 1 = tan^2A, 1/cos^2A = sec^2A
sec^2A - 1 = tan^2A, This is a trig identity

sinA/cosA * (1 - cot^2A) + cosA/sinA * (1 - tan^2A) = 0
sinA/cosA = tanA, cosA/sinA = cotA: tanA * (1-cot^2A) + cotA * (1-tan^2A) = 0
Multiply through: tanA - cotA + cotA - tanA = 0, You can see that the remaining terms will cancel out, leaving you with 0 = 0

sinA * cosA * (tanA + cotA) = 1, remember that tanA = sinA/cosA and cotA = cosA/sinA
Multiply through: sinA*cosA*tanA + sinA*cosA*cotA = 1
cosA cancels in first term, while sinA cancels in second term: sin^2A + cos^2A = 1, this is another trig identity