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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.

Update:

Thank you both!! You're awesome.

5 Answers

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  • Favorite 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

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  • 8 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

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  • Ashley
    Lv 6
    8 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

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  • Anonymous
    8 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

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  • 8 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

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