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Thread: problem with proving identities using basic identities

  1. #1
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    problem with proving identities using basic identities

    i have a problem with this:

    prove,
    a)$\displaystyle \frac{1-tan^2\theta}{cot^2 \theta-1}=tan^2\theta$


    b)$\displaystyle (cos \theta+sin \theta)^2+(cos \theta-sin \theta)^2=2$

    any help will appreciate!! thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    a) $\displaystyle \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta }{sin^2\theta} - 1}$

    $\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

    $\displaystyle =\frac{\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta }}$

    $\displaystyle = \frac{sin^2\theta}{cos^2\theta}$

    $\displaystyle =tan^2\theta$
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  3. #3
    MHF Contributor Unknown008's Avatar
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    b) $\displaystyle (cos\theta + sin\theta)^2 + (cos\theta - sin\theta)^2 = cos^2 \theta + 2cos\theta sin\theta + sin^2\theta + cos^2\theta -2cos\theta sin\theta + sin^2\theta$

    $\displaystyle = cos^2 \theta + \cancel{2cos\theta sin\theta} + sin^2\theta + cos^2\theta \cancel{-2cos\theta sin\theta} + sin^2\theta$

    $\displaystyle = 2cos^2\theta + 2 sin^2\theta$

    $\displaystyle = 2(cos^2\theta + sin^2\theta)$

    $\displaystyle = 2(1)$

    $\displaystyle = 2$
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  4. #4
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    Quote Originally Posted by Unknown008 View Post
    a) $\displaystyle \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta }{sin^2\theta} - 1}$

    $\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

    $\displaystyle =\frac{\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta }}$

    $\displaystyle = \frac{sin^2\theta}{cos^2\theta}$

    $\displaystyle =tan^2\theta$

    i want to know,how can $\displaystyle \frac{cos^2\theta-sin^2\theta}{cos^2 \theta}=\frac{1}{cos^2\theta}$ ??
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  5. #5
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    Because $\displaystyle \frac{\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta}}}{\frac{\cos^2{\the ta} - \sin^2{\theta}}{\sin^2{\theta}}} = \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t heta}}}\left(\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}\right)$

    $\displaystyle = \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t heta}}}$
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, I'll break up the steps, but this is not what I meant.

    $\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

    $\displaystyle = \frac{cos^2\theta - sin^2\theta}{cos^2\theta} \times \frac{sin^2\theta}{cos^2\theta - sin^2\theta}$

    Then:

    $\displaystyle = \frac{1}{cos^2\theta} \times \frac{sin^2\theta}{1}$
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  7. #7
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    thanks very much both of you sir!!
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