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prove: (cosecA - sinA)(SecA - CosA) = 1/CotA + TanA

by Guest6012  |  12 years, 9 month(s) ago

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prove: (cosecA - sinA)(SecA - CosA) = 1/CotA + TanA

 Tags: 1CotA, Cosa, cosecA, prove, sinASecA, tanA

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  1. Guest2038
    LHS: (sinA - cosA + 1) / (sinA + cosA - 1)
    Multiply Nr and Dr by cosA:
    Nr: cosA(sinA - cosA + 1)
    Dr: CosA(sinA + cosA + 1)

    Nr = cosAsinA - cos²A + cosA = cosAsinA -(1-sin²A) + cosA
    = cosAsinA + cosA + sin²A - 1 (Rearranging)
    = cosA(sinA + 1) + (sinA+1) (sinA - 1) [since a² - b² = (a+b)(a-b)]
    = (sinA+1) (cosA + sinA -1)
    Now Nr/Dr gives us:
    = [(sinA+1)(sinA + cosA - 1)] / [cosA(sinA + cosA - 1)]
    = {sinA+1} / cosA (After cancelling the like term sinA+cosA+1 in both the Nr and Dr)
    = {sinA/cosA} + {1/cosA} = tanA + secA = RHS

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