Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

Prove this trig identity?

(secA-cosA)/(cosA-sinA)

=

sinA/cotAcosA-cosA

2 Answers

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  • TomV
    Lv 7
    2 years ago
    Favorite Answer

    LHS = (secA-cosA)/(cosA-sinA)

    = (1 - cos²A)/(cos²A - cosAsinA) : multiplication of denominator and numerator by cosA ≠ 0

    = sin²A/(cos²a - cosAsinA) : application of Pythagorean Theorem, 1 - cos²A = sin²A

    = sinA/(cos²A/sinA - cosA) : divide denominator and numerator by sinA ≠ 0

    = sinA/(cotAcosA - cosA) : application of definition of cotA = cosA/sinA

    = RHS

  • 2 years ago

    69

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