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Thread: tan^2(x)-sin^2(x)

  1. #1
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    Question tan^2(x)-sin^2(x)

    Question asks to find equivalent of $\displaystyle \tan ^2 (x) - \sin ^2 (x)$

    with selection of A, B, C, or D below respectively
    $\displaystyle
    \begin{array}{l}
    \sec ^2 (x) \\
    2\cot (x)\csc (x) \\
    \frac{{\sin ^3 (x)}}{{\cos ^2 (x)}} \\
    \tan ^2 (x)\sin ^2 (x) \\
    \end{array}
    $

    I've made several attempts at this and still draw blank, below is my latest.

    $\displaystyle \frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - \sin ^2 (x) \\ $

    $\displaystyle \frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - (1 - \cos ^2 (x)) \\ $

    $\displaystyle \frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - 1 + \cos ^2 (x) \\ $

    $\displaystyle \frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - \frac{{\cos ^2 (x)}}{{\cos ^2 (x)}} + \frac{{\cos ^4 (x)}}{{\cos ^2 (x)}} \\ $

    $\displaystyle \frac{{\sin ^2 (x) - \cos ^2 (x) + \cos ^4 (x)}}{{\cos ^2 (x)}} \\ $

    $\displaystyle \frac{{(1 - \cos ^2 (x)) - \cos ^2 (x) + \cos ^4 (x)}}{{\cos ^2 (x)}} \\ $

    $\displaystyle \frac{{1 - 2\cos ^2 (x) + \cos ^4 }}{{\cos ^2 (x)}} \\ $

    And here a come to what appear to be a dead end
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    $\displaystyle \begin{aligned}
    \tan ^{2}(x)-\sin ^{2}(x)&=\sin ^{2}(x)\left( \frac{1}{\cos ^{2}(x)}-1 \right) \\
    & =\sin ^{2}(x)\cdot \frac{\overbrace{1-\cos ^{2}(x)}^{\sin ^{2}(x)}}{\cos ^{2}x} \\
    & =\overbrace{\frac{\sin ^{2}(x)}{\cos ^{2}(x)}}^{\tan ^{2}(x)}\cdot \sin ^{2}(x) \\
    & =\tan ^{2}(x)\sin ^{2}(x).
    \end{aligned}$
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  3. #3
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    Thankyou for your reply. I'm not sure how you got to the first step though
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    I just factorised.
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  5. #5
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    Thumbs up

    Oh okay now I see it. Thankyou.
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