# Pre-calculus identities help?

I have a take home test I'm trying to do and I have most of it done but I need some help with these.

These are establishing identities

0=theta

1. Cos0(1 + tan^2 0) = 1

2. (sin0)/(1+cos0) + (1 + cos0)/(sin0) = 2csc0

These are sum and difference I think. There are a few of these but I hopefully I can figure the rest out.
a= alpha B = beta

Find sin (a + B), cos (a + B), sin (2a), cos(2a), sin (B/2)

3. Tan(a) = -2 Pi/2 < a < pi
CotB = -2. Pi/2 < B < pi

0 = θ

1.

cosθ(1 + tan2θ)

Recall : Trigonometry sec2θ - tan2θ = 1

= cosθ(sec2θ - tan2θ + tan2θ)

= cosθ(sec2θ)

Recall : Trigonometry formulas sec2θ = 1 / cos2θ

= cosθ(1 / cos2θ)

=1 / cosθ

=secθ

But θ = 0

=sec0

= 1.

0 = θ

2.

sinθ / (1 + cosθ) + (1 + cosθ) / sinθ

= sinθ(1 - cosθ) / (1 + cosθ)(1 - cosθ) + (1 + cosθ) / sinθ

= sinθ(1 - cosθ) / (1 - cos2θ) + (1 + cosθ) / sinθ

Recall : Trigonometry identities sin2θ + cos2θ = 1

= sinθ(1 - cosθ) / (sin2θ) + (1 + cosθ) / sinθ

= (1 - cosθ) / sinθ + (1 + cosθ) / sinθ

= (1 - cosθ + 1 + cosθ) / sinθ

= 2 / sinθ

Recall : Trigonometry formulas 1 / sinθ = cosθ

= 2cscθ.

sin(a + B)

Recall : Trigonometry identities  for sum of two angles Sin (A + B) = sinAcosB + cosAsinB

Substitute A = a and B = B in the  above identities

sin(a + B) = sinacosB + cosasinB.

cos(a + B)

Recall : Trigonometry identities  for sum of two angles cos (A + B) = cosAcosB - sinAsinB

Substitute A = a and B = B in the  above identities

cos(a + B) = cosacosB - sinasinB.

sin(2a)

Recall : Trigonometry identities  for sum of two  equal angles sin (2A) = 2sinAcosA

Substitute A = a  in the  above identities

sin(2a) = 2sinacosa.

cos(2a)

Recall : Trigonometry identities  for sum of two  equal angles cos(2A) = 2cos2A - 1

Substitute A = a  in the  above identities

cos(2a) = 2cos2a - 1

=2(1 - sin2a) - 1

=2 - 2sin2a - 1

=1 - 2sin2a

=sin2a + cos2a - 2sin2a

=cos2a - sin2a.

sin(B / 2)

Recall : Trigonometry identities  for one angle sin(A / 2) =±√[(1 - cosA) / 2]

Substitute A = B  in the  above identities

sin(B / 2) = ±√[(1 - cosB) / 2].

3.

Tan(a) = -2    π/2 < a < π   -(1)

CotB = -2      π/2 < B < π   -(2)

From the first and second equations we get

Tan(a) = cotB

Tan(a) = Tan(3π/2 - B)

Remove Tan to each side

a = 3π/2 - B

a + B = 3π/2

sin(a + B) = sin(3π/2)

= sin(π + π/2)

= - sin(π/2)

= - 1.

3.

Tan(a) = -2    π/2 < a < π   -(1)

CotB = -2      π/2 < B < π   -(2)

From the first and second equations we get

Tan(a) = cotB

Tan(a) = Tan(3π/2 - B)

Remove Tan to each side

a = 3π/2 - B

a + B = 3π/2

cos(a + B) = cos(3π/2)

= cos(π + π/2)

= - cos(π/2)

= 0.

3.

Tan(a) = -2    π/2 < a < π   -(1)

CotB = -2      π/2 < B < π   -(2)

Recall : Trigonometry identities sec^2a = 1 + tan^2a

Sec^2(a) = 1 + (-2)^2

= 1 + 4

= 5

sec^a = √5

cosa = ±(1 / √5)

sina = √( 1 - cos^2a)

= √(1 - (1 / 5)

= √4 / 5

= ±(2 / √5)

sin(2a) = 2sinacosa

=2(2 / √5)(1 / √5)

=4 / 5.