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Math Help - Prove an identity and solve an equation

  1. #1
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    Prove an identity and solve an equation

    Prove the following identity:

    [tan(x)+cot(x)]/[sec(x)-tan(x)] = sec^2(x) + sec(x)tan(x)csc^2(x)


    Solve the Following Expression:

    (16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

    I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!
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  2. #2
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    Hello jhunt47
    Quote Originally Posted by jhunt47 View Post
    Prove the following identity:

    [tan(x)+cot(x)]/[sec(x)-tan(x)] = sec^2(x) + sec(x)tan(x)csc^2(x)


    Solve the Following Expression:

    (16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

    I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!
    \tan x + \cot x = \frac{\sin x}{\cos x}+ \frac{\cos x}{\sin x}
    =\frac{\sin^2x+\cos^2x}{\sin x\cos x}

    =\frac{1}{\sin x\cos x}
    and \sec x - \tan x = \frac{1}{\cos x}-\frac{\sin x}{\cos x}
    =\frac{1-\sin x}{\cos x}
    So \frac{\tan x + \cot x}{\sec x - \tan x}=\frac{1}{\sin x \cos x}\div\frac{1-\sin x}{\cos x}
    =\frac{1}{\sin x \cos x}\times\frac{\cos x}{1-\sin x}

    =\frac{1}{\sin x(1-\sin x)}

    =\frac{1+\sin x}{\sin x(1-\sin x)(1+\sin x)}

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

    =\frac{1+\sin x}{\sin x\cos^2x}

    =\frac{1}{\sin x \cos^2x}+\frac{1}{\cos^2x}

    =\frac{\sin x}{\sin^2 x \cos^2x}+\sec^2x

    =\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{\sin^2x}+\sec^2 x

    =\sec x \tan x \csc^2 x + \sec^2 x
    No time now to look at part 2. I'll do so later if no-one else has done it in the meantime.

    Grandad
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  3. #3
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    Hello again jhunt47

    This is an interesting one!
    Quote Originally Posted by jhunt47 View Post
    Solve the Following Expression:

    (16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

    I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!
    First notice that
    \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot\left(\fra  c{16}{81}\right)^{-\sin^2x}
    So, if we let y = \left(\frac{16}{81}\right)^{\sin^2x}, then
    \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot y^{-1}=\left(\frac{16}{81}\right)\cdot\frac{1}{y}=\fra  c{16}{81y}
    and the equation becomes
    y+\frac{16}{81y}=\frac{26}{27}

    \Rightarrow 81y^2+16=78y

    \Rightarrow 81y^2-78y+16=0

    \Rightarrow (3y-2)(27y-8)=0

    \Rightarrow y = \left(\frac{16}{81}\right)^{\sin^2x}= \frac23, \,\frac{8}{27}
    We now note that \left(\frac{16}{81}\right)^{\frac14}=\frac23 and \left(\frac{16}{81}\right)^{\frac34}=\frac{3}{27}
    \Rightarrow\sin^2x=\frac14,\,\frac34

    \Rightarrow \sin x = \pm\frac12,\,\pm\frac{\sqrt3}{2}

    \Rightarrow x = n\pi \pm \frac{\pi}{6} or n\pi \pm \frac{\pi}{3},\;n \in \mathbb{Z}
    Grandad
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