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I am not sure if you meant the right hand side to be cosecant or cosine. I will do both.

1 + sinΘ = cosΘ

1 = cos Θ - sin Θ

Square both sides:

1 = cos^2(Θ) +sin^2(Θ) - 2sinΘcosΘ

1 = 1 - 2sinΘcosΘ

0 = 2sinΘcosΘ

0 = sin 2Θ

2Θ = 0, pi, 2pi, 3pi, 4pi, ...

Θ = 0, (pi/2), pi, 3pi/2

Θ = 0, 90, 180, 270

Evaluating these solutions into the original equation you get that Θ = 0 and 270 solve the original equation.

1 + sinΘ = cscΘ

1 + sinΘ = 1/ sinΘ

Multiply both sides by sinΘ

sinΘ + sin^2(Θ) = 1

sin^2(Θ) + sinΘ - 1 = 0

This is a quadratic in sinΘ so use the quadratic formula

sinΘ = (1/2)[ -1 +/- sqrt(1 +4)]

sinΘ = (1/2)[-1 + sqrt(5)]--> -1 - sqrt(5) outside of domain

Θ ~ 38.17

1 + sin(θ) = cos(θ)

1 + sin(θ) = √(1 - sin^2(θ))

( -1 <= sin(θ) <= 1, so that means 0 <= cos(θ) <= 2, so the right hand side has to be positive and we don't have to worry about a ±)

(1 + sin(θ))^2 = 1 - sin^2(θ)

1 + 2sin(θ) + sin^2(θ) = 1 - sin^2(θ)

2sin(θ) + 2sin^2(θ) = 0

sin(θ) + sin^2(θ) = 0

sin(θ)(1 + sin(θ)) = 0

So θ = 0 or 180; or 270

But checking with the original problem, 180 doesn't work. So the only answers are 0 and 270.

1 + sin = cos

(1+ sin)² = cos²

1 + 2sin + sin² = cos²

note: cos² = 1 - sin², so

1 +2sin +sin² = 1 - sin²

2sin² +2sin = 0

2sin (sin + 1) = 0

sin (sin+1) = 0

sin = 0 yields angle = 0 degrees

sin+1 = 0

sin = -1 yields angle = 270 degrees

270