Anonymous
Anonymous asked in Education & ReferenceHomework Help · 9 years ago

I need help proving the following trigonometric function:?

(secA + 1 )/(sin A + tan A) = (1)/(sinA)

I have to show all steps and only alter one side to make it equal to the other and I cannot do this problem. Help would be very appreciated thank you

3 Answers

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  • Anonymous
    9 years ago
    Favorite Answer

    * LHS *

    Note that tan(A) = sin(A)/cos(A) and sec(A) = 1/cos(A). Then..

    (1/cos(A) + 1)/(sin(A) + sin(A)/cos(A)) * cos(A)/cos(A)

    = (1 + cos(A))/(sin(A)cos(A) + sin(A))

    Factoring the bottom expression and reducing the common factors of 1 + cos(A), we have..

    (1 + cos(A))/((1 + cos(A))sin(A))

    = 1/sin(A)

    = RHS

    I hope this helps!

    Source(s): Knowledge
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  • vohs
    Lv 4
    4 years ago

    Wow you're rather on a roll. Asking a number of those questions approximately trigonometry. besides this one is trouble-free. Mulitply the denominator and the numerator contained in the best hand element with (a million + Cos B). The denominator will become ( a million - Cos^2 B) = Sin^2 B. Now merely divide the two Sin^3 B by utilising Sin ^2 B and multiply it with ( a million + Cos B).

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  • 9 years ago

    (secA + 1)/(sinA + tanA) (write secant and tangent in terms of sine and cosine)

    (1/cosA + 1)/(sinA + sinA/cosA) (multiply the numerator and denominator by cosA)

    (1 + cosA)/(cosAsinA + sinA) (factor out sinA from the denominator)

    (1 + cosA) / sinA(cosA + 1) (cancel out 1 + cosA from the numerator and denominator)

    1/sinA

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