Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

Help with trig identities?

Im stuck on this one..

(secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA)

=

4tanAcscA

??

4 Answers

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  • 2 years ago
    Favorite Answer

    OK, here you go sweetheart :)

    Note the identities used in the proof :

    secA = 1/cosA

    tanA = sinA/cosA

    1 - sin²A = cos²A

    Starting with the LHS we have :

    ∴ LHS

    = [(secA + tanA) /(secA - tanA)] - [(1 - sinA) /(1 + sinA)

    = [(1/cosA + sinA/cosA) /(1/cosA - sinA/cosA)] - [(1 - sinA) /(1 + sinA)]

    = [[(1 + sinA) /cosA] ÷ [(1 - sinA) /cosA]] - [(1 - sinA) /(1 + sinA)]

    = [(1 + sinA) /(1 - sinA)] - [(1 - sinA) /(1 + sinA)]

    = [(1 + sinA)² - (1 - sinA)²] /(1 - sin²A)

    = [1 + 2sinA + sin²A - 1 + 2sinA - sin²A] /(1 - sin²A)

    = [4sinA] /cos²A

    = 4 (sinA/cosA) (1/cosA)

    = 4(tanA)(secA)

    = RHS, as required, Q.E.D.

    Oh, I forgot to mention that what you've written in your question is NOT an identity! It should be a 'secA' instead of a 'cosecA' at the end for the identity to be valid.

    Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    P.S. (Don't forget to vote me best answer as being the first to correctly and thoroughly answer your question!)

    • TomV
      Lv 7
      2 years agoReport

      You didn't really answer the question, 'though. The RHS of the posted identity is 4tanAcscA, not 4 tanAsecA.

  • Pope
    Lv 7
    2 years ago

    Take it or leave it, but on these identity proofs I always begin by checking whether there are any conflicts in the domains of the two sides of the equation. In this case there are.

    Let A = 0.

    The left side of the equation is zero. The right side is undefined. When a real number is equated with an undefined expression, the equation cannot be an identity. And no, there is no such thing as expressions that are almost an identical.

  • TomV
    Lv 7
    2 years ago

    The reason you are stuck on that one may be because it is not an identity.

    Let A = 60°

    sinA = √3/2

    secA = 1/cosA = 2

    cscA = 1/sinA = 2/√3

    tanA = √3

    (secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA) = 4tanAcscA

    (2 + √3)/(2 - √3) - (1-√3/2)/(1+√3/2) = 4(√3)(2/√3)

    (2+√3)/(2-√3) - (2-√3)/(2+√3) = 8

    (2+√3)²/(4-3) - (2-√3)²/(4-3) = 8

    4 + 4√3 + 3 - (4 - 4√3 + 3) = 8

    7 + 4√3 - 7 + 4√3 = 8

    8√3 = 8 : False statement

    Equation is not an identity.

    However

    (secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA) = 4tanAsecA

    is an identity. (Note the different RHS)

    LHS = (secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA)

    replace secA with 1/cosA and tanA with sinA/cosA

    = (1/cosA + sinA/cosA)/(1/cosA - sinA/cosA) - (1-sinA)/(1+sinA)

    multiply numerator and denominator of first term by cosA

    = (1+sinA)/(1-sinA) - (1-sinA)/(1+sinA)

    multiply numerator and denominator of first term by (1+sinA) and second term by (1-sinA)

    = (1+sinA)²/(1-sin²A) - (1-sinA)²/(1 - sin²A)

    replace 1 - sin²A with cos²A per Pythagorean Identity, and combine terms with common denominator

    = [(1+sinA)² - (1-sinA)²]/cos²A

    expand squared terms

    = [(1 + 2sinA + sin²A - (1 - 2sinA + sin²A)]/cos²A

    combine like terms

    = 4sinA/cos²A

    replace sinA/cosA with tanA and 1/cosA with secA

    = 4tanAsecA

    = RHS (as modified)

  • 2 years ago

    Try multiplying the fractions by their conjugate (sec A + tan A) or (sec A - tan A) and (1 - sin A) or (1 + sin A) to create identities (sec²A - tan²A = 1) and (1 - sin²A = cos²A)

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