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Math Help - Prove this

  1. #1
    Member kjchauhan's Avatar
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    Prove this

    Pl Prove this:

    \tan^{-1}(\cosh\theta - \sinh\theta) = \frac{1}{2}\sin^{-1}(sech\theta)

    Thanks in advance.
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  2. #2
    Junior Member
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    Quote Originally Posted by kjchauhan View Post
    Pl Prove this:

    \tan^{-1}(\cosh\theta - \sinh\theta) = \frac{1}{2}

    Thanks in advance.
    Rewrite the identity:
    2\tan^{-1}(\cosh\theta - \sinh\theta) = \sin^{-1}(sech\theta)

    Take tan of both sides:

    \tan{(2\tan^{-1}(\cosh\theta - \sinh\theta))} = tan(\sin^{-1}(sech\theta))

    \cosh\theta=\frac{1}{2}(e^x+e^{-x})

    \sinh\theta=\frac{1}{2}(e^x-e^{-x})

    Hence:
    \cosh\theta - \sinh\theta=e^{-x}

    Our identity becomes:

    \tan{(2\tan^{-1}{e^{-x}})} = tan(\sin^{-1}(sech\theta)

    Now we can work with LHS only - by using the double angle formula:

    \tan{2A}=\frac{2tanA}{1-tan^2A}
    in our case A=\tan^{-1}{e^{-x}}

    Hence LHS= \frac{2e^{-x}}{1-e^{-2x}}=\frac{2}{e^x-e^{-x}}<br />

    Now we simplify the RHS

    RHS=\tan(\sin^{-1}(sech\theta)

    =\frac{\sin(\sin^{-1}(sech\theta)}{\cos(\sin^{-1}(sech\theta))}

    =\frac{sech\theta}{\sqrt{1-\sin^2{\sin^{-1}{sech\theta}}}}

    =\frac{sech\theta}{\sqrt{1-sech^2\theta}}

    =\frac{sech\theta}{\sqrt{1-\frac{1}{\cosh^2\theta}}}

    =\frac{sech\theta cosh\theta}{\sqrt{\cosh^2\theta-1}}

    Use the identity:

    \cosh^2\theta-\sinh^2\theta=1

    Hence:

    \cosh^2\theta-1=\sinh^2\theta

    So RHS becomes:

    =\frac{sech\theta cosh\theta}{\sqrt{\sinh^2\theta}}

    =\frac{\frac{1}{cosh \theta}cosh \theta}{sinh\theta}

    =\frac{1}{sinh \theta}

    =\frac{2}{e^x-e^{-x}} which is the LHS
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