trigonometry

find acute angle A and B satisfying secAcotB-secA-2cotB+2=0

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  1. secAcotB-secA-2cotB+2=0
    (1/cosA)(cosB/sinB) - (1/cosA)- 2(cosB/sinB) = -2
    multiply each term by sinBcosA
    cosB - sinB - 2cosBcosA = -2cosAsinB
    cosB - sinB - 2cosBcosA + 2cosAsinB = 0
    (cosB - sinB) - 2cosA(cosB - sinB) = 0
    (cosB - sinB)(1 - 2cosA) = 0

    cosB = sinB OR 1 - 2cosA = 0

    if cosB = sinB
    sinB/cosB = 1
    tanB = 1
    B = 45° or π/4

    if 1-2cosA = 0
    cosA = 1/2
    A = 60° or π/3

    so A=60° , B= 45°
    A=π/3 , B=π/4

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    2. 👎 1
  2. secAcotB-secA-2cotB+2=0
    (1/cosA)(cosB/sinB) - (1/cosA)- 2(cosB/sinB) = -2
    multiply each term by sinBcosA
    cosB - sinB - 2cosBcosA = -2cosAsinB
    cosB - sinB - 2cosBcosA + 2cosAsinB = 0
    (cosB - sinB) - 2cosA(cosB - sinB) = 0
    (cosB - sinB)(1 - 2cosA) = 0

    cosB = sinB OR 1 - 2cosA = 0

    if cosB = sinB
    sinB/cosB = 1
    tanB = 1
    B = 45° or π/4

    if 1-2cosA = 0
    cosA = 1/2
    A = 60° or π/3

    so A=60° , B= 45°

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  3. Sin

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  4. Thanks

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  5. Thanks a lot. It helped me while practising for exam.

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  6. Thnx Tammy om and all of u posted this sum lots of thnx

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    2. 👎 3

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