maths

Prove that
[1+cotA-CosecA] [1+tanA+secA] = 2

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  3. 👁 724
  1. [1+cotA-CosecA]*[1+tanA+secA]
    = 1 + tanA + secA + cotA + 1 + cosecA - (1/sinA) - (1/cosA) - secAcosecA
    = 1 + tanA + secA + cotA + 1 + cosecA - cosecA - secA - secAcosecA
    = 2 + tanA + cotA - secAcosecA
    = 2 + [(sin^2A + cos^2A)/(sinAcosA)] - secAcosecA
    = 2 + 1/(sinAcosA) - secAcosecA
    = 2 + secAcosecA - secAcosecA
    = 2

    1. 👍 8
    2. 👎 2
  2. 1+cot-csc = (sin+cos-1)/sin
    1+tan+sec = (sin+cos+1)/cos
    multiply them and you have
    ((sin+cos)^2-1)/(sin*cos)
    = (1+2sin*cos-1)/(sin*cos)
    = 2

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    2. 👎 0

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