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Math Help - 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 \tan ^2 (x) - \sin ^2 (x)

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

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

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

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

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

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

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

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

     \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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    \begin{aligned}<br />
   \tan ^{2}(x)-\sin ^{2}(x)&=\sin ^{2}(x)\left( \frac{1}{\cos ^{2}(x)}-1 \right) \\ <br />
 & =\sin ^{2}(x)\cdot \frac{\overbrace{1-\cos ^{2}(x)}^{\sin ^{2}(x)}}{\cos ^{2}x} \\ <br />
 & =\overbrace{\frac{\sin ^{2}(x)}{\cos ^{2}(x)}}^{\tan ^{2}(x)}\cdot \sin ^{2}(x) \\ <br />
 & =\tan ^{2}(x)\sin ^{2}(x).<br />
\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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