Help Please...Trig Proofs?

Question

I'm really having trouble with these two proofs. Can anyone solve these two? It would be greatly appreciated! I'm having trouble getting one side to equal the other.

1.) (sinθ -1)(tanθ + secθ ) = -cosθ

2.) cos B cot B = csc B - sin B

Answer 1

(I don't know how to type thetas, so I'm using x's)

1.) (sinθ -1)(tanθ + secθ )
= sinx tanx + sinx secx - tanx - secx
= (sin^2x / cosx) + (sinx / cosx) - (sinx / cosx) - (1 / cosx)
= (sin^2x - 1) / cosx
= -cos^2x / cosx
= -cosx

2.) cos B cot B
= cos B (cosB / sin B)
= cos^2 B / sin B
= (1 - sin^2 B) / sin B
= (1 / sin B) - (sin^2 B / sin B)
= csc B - sin B

Answer 2

Question 1
(sin θ - 1)(sin θ/cos θ + 1/cos θ)
(sin θ - 1)(sinθ + 1) / (cos θ)
(sin ² θ - 1) / cos θ
- (1 - sin ² θ) / cos θ
- cos ² θ / cos θ
- cos θ

Question 2
cos B cos B / sin B
cos ² B / sin B
(1 - sin ² B) / sin B
cosec B - sin B

Answer 3

1. secT = 1/cosT. tanT = sinT/cosT
(sinT - 1) (sinT/cosT + 1/cosT) = -cosT ?
(sinT - 1)( (sinT + 1)/cosT ) = -cosT ?
(sin²T + sinT - sinT - 1)/cos T = -cosT ?
[from relationship sin²T + cos²T = 1, sin²T - 1 = -cos²T ]
-cos²T/cosT = -cosT ?
-cosT = -cosT qed.

2. cotB = cosB/sinB; cscB = 1/sinB
cosBcotB = cscB - sin B ?
cosB(cosB/sinB) = 1/sinB - sinB ?
cos²B/sinB = (1-sin²B)/sinB ?
(1-sin²B)/sinB = (1-sin²B)/sinB qed.

If you don't know sin²B + cos²B = 1, take a few minutes to prove it to yourself.