# Find all solutions for: sec β csc β = 2csc β?

2. Find all solutions for: tan ϴ + sec ϴ= 1

*Please show these step-by-step. I would really appreciate it.*

Thank You in Advance.

Thank you both!! You're awesome.

### 5 Answers

- 8 years agoFavorite Answer
sec(b)csc(b) = 2csc(b)

sec(b)csc(b) - 2csc(b) = 0

csc(b) * (sec(b) - 2) = 0

csc(b) = 0

1/sin(b) = 0

No solution

sec(b) - 2 = 0

sec(b) = 2

1/cos(b) = 2

cos(b) = 1/2

b = pi/3 + 2pi * k , 5pi/3 + 2pi * k

b = (pi/3) * (1 + 6k) , (pi/3) * (5 + 6k)

tan(t) + sec(t) = 1

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

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

sin(t) + 1 = cos(t)

(sin(t) + 1)^2 = cos(t)^2

sin(t)^2 + 2sin(t) + 1 = cos(t)^2

sin(t)^2 + 2sin(t) + 1 = 1 - sin(t)^2

2sin(t)^2 + 2sin(t) = 0

sin(t) * (sin(t) + 1) = 0

sin(t) = 0

t = pi * k

sin(t) + 1 = 0

sin(t) = -1

t = 3pi/2 + 2pi * k

t = (pi/2) * (3 + 4k)

Test each answer:

t = 0 , pi , 3pi/2

tan(0) + sec(0) =>

0 + 1 =>

1

tan(pi) + sec(pi) =>

0 + (-1) =>

-1

tan(3pi/2) + sec(3pi/2) =>

Does not exist

t = 0 , 2pi , 4pi , ...

t = 2pi * k, k is an integer, is the answer

- Login to reply the answers

- peabodyLv 78 years ago
sec β csc β = 2csc β

1/ cos β * 1/ sin β = 2 / sin β

1/ sin β cos β = 2 / sin β

cross multiply

sin β = sin β cos β

sin β cos β - sin β = 0

sin β ( cos β -1) =0

either sin β = 0

β = sin^-1 (0) = 0 , pi , 2pi for each cycle

OR cos β = 1

β = cos^-1 (1) = 0 , 2pi for each cycle

So β = npi where n is an integer

- Login to reply the answers

- AshleyLv 68 years ago
1. sec β csc β = 2csc β divide both sides by cscβ

sec β csc β = 2csc β

......csc β.....= 2

cscβ cancels to get Secβ = 2

or cosβ = ½

Therefore β = 2π (n ± 1/3 where n is an integer

Change to sin&cos .........sinϴ/cosϴ + 1/cosϴ

Write with common denominator:...(sinϴ + 1) / cosϴ = 1

Cross Multiply: .................sinϴ + 1 = cosϴ

Square both sides ..........Sin²ϴ + 2sinϴ + 1 = cos²ϴ

Pyth identity ................ Sin²ϴ + 2sinϴ + 1 = 1 - sin²ϴ

All on one side ...........2 sin²ϴ + 2sinϴ = 0

Factor ................2sinϴ(sinϴ+1) = 0

so sinϴ = 0 or sinϴ = -1

ϴ = 2nπ or 3π /2 ± 2nπ

- Login to reply the answers

- Anonymous8 years ago
1/

equivalent to sec b = 2 ---> cos b = 1/2 ---> b = pi/3 , 5pi/3

2/

sin t + 1 = cos t ---> cost - sint = 1 --> (sqrt2) cos(t + (pi/4)) = 1 , now you can solve

- Login to reply the answers

- How do you think about the answers? You can sign in to vote the answer.
- Big OrangeLv 68 years ago
divide both sides of csc(β)

sec(β) = 2

(1/cos(β)) = 2

cos(β) = 1/2

β = 60 degrees = 300 degrees

tan(ϴ) + sec(ϴ) = 1

sin(ϴ)/cos(ϴ) + 1 / cos(ϴ) = 1

(sin(ϴ)+1) / cos(ϴ) = 1

sin(ϴ) + 1 = cos(ϴ)

1 =cos(ϴ) - sin(ϴ)...square both sides

1 =(cos(ϴ) - sin(ϴ))^2...multiply it out and simplify using identities to get

1 = 1-2sin(ϴ)cos(ϴ)

0 = -2sin(ϴ)cos(ϴ)

0=sin(ϴ)cos(ϴ)

sin(ϴ)=0.........ϴ = 0, 180, 360

cos(ϴ)=0........ϴ = 90,270

- Login to reply the answers