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Math Help - Trig Identities Help

  1. #1
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    Trig Identities Help

    Hi guys, im doing a Maths quiz, and have gone through the basic Trig Identity stuff, but I am stuck on 2 questions....Anyone give me an idea what to do?

    Which of the following is equal to (1 − cos2y)(1 + cot2y)

    A) cos(2 y) B) 1 C) sec^2y D) sin^2y

    and

    Which of the following is equal to

    1/ (sin x X cos x) − (cos x /sin x)


    A) − 1 / (sin(2 x))

    B) cos^2x

    C) 1

    D tan x

    E) 1/ sin(2 x)

    F) None of the above
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  2. #2
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    Okay, I got the answer of 1 for the first question..

    Stuck on the second though..

    Any ideas?
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, Maccabhoy!

    Here's the second problem . . .


    Which of the following is equal to: . \frac{1}{\sin x\cos x} - \frac{\cos x}{\sin x}

    . . (A)\;\frac{-1}{\sin2x} \quad(B)\;\cos^2\!x \qquad (C)\;1\qquad (D)\;\tan x \qquad <br />
(E)\;\frac{1}{\sin2x} \qquad (F)\text{ None of these}
    \text{We have: }\;\frac{1}{\sin x\cos x} - \frac{\cos x}{\sin x}\cdot{\color{blue}\frac{\cos x}{\cos x}} \;\;=\;\;\frac{\overbrace{1 - \cos^2\!x}^{\text{This is }\sin^2\!x}}{\sin x\cos x}


    . . . . . . = \;\frac{\sin^2\!x}{\sin x\cos x} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x . . . answer (D)

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