Verify the identities algebraically? (Trig)?

Question

For some reason I cannot figure out these three problems. If someone would help me and go over it with me I would greatly appreciate it.(0=theta)

1+csc0/sec0-cot0=cos0

sin0/1-cos0=1+cos0/sin0

cos^2(0)-sin^2(0)=2cos^2(0)-1

Answer 1

2) sin(t)/[1 - cos(t)] = [1 + cos(t)]/sin(t)

LHS = sin(t)/[1 - cos(t)]

Multiply top and bottom by the bottom's conjugate.

LHS = sin(t)[1 + cos(t)] / [ (1 - cos(t))(1 + cos(t)) ]

The bottom becomes a difference of squares.

LHS = sin(t)[1 + cos(t)] / [ 1 - cos^2(t) ]

Use the identity that 1 - cos^2(t) = sin^2(t).

LHS = sin(t)[1 + cos(t)] / sin^2(t)

Cancel the single sin(t) on the top with one of the sin(t) on the bottom.

LHS = (1 + cos(t))/sin(t) = RHS

3) cos^2(t) - sin^2(t) = 2cos^2(t) - 1

LHS = cos^2(t) - sin^2(t)

Use the identity that sin^2(t) = 1 - cos^2(t).

LHS = cos^2(t) - [1 - cos^2(t)]
LHS = cos^2(t) - 1 + cos^2(t)

Group like terms.

LHS = 2cos^2(t) - 1 = RHS

Your first question is confusing to interpret. I ask that you please use bracketing to differentiate between top and bottom of the denominator (reading it is ambiguous).

For instance, if somebody wanted to write
(x + 3)/(x^2 + 6x + 9) and wrote it as
x + 3/x^2 + 6x + 9, it wouldn't be clear due to order of operations.

Answer 2

Use the identities in the link below.

(I'll use x instead of theta)
e.g., Sin^2(x) + Cos^2(x) = 1
then Sin^2(x) = 1 - Cos^2(x)
(which we found with a very simple algebraic move)

Take your third equation above:
cos^2(x)-sin^2(x)=2cos^2(x)-1

Let's take the left side:
Cos^2(x) - Sin^2(x)
Cos^2(x) - (1 - Cos^2(x))
Cos^2(x) - 1 + Cos^2(x)
2 Cos^2(x) - 1

Done