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  1. #1
    Member kjchauhan's Avatar
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    Prove this

    Pl Prove this:

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

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

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

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

    Take tan of both sides:

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

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

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

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

    Our identity becomes:

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

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

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

    Hence LHS=$\displaystyle \frac{2e^{-x}}{1-e^{-2x}}=\frac{2}{e^x-e^{-x}}
    $

    Now we simplify the RHS

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

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

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

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

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

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

    Use the identity:

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

    Hence:

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

    So RHS becomes:

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

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

    $\displaystyle =\frac{1}{sinh \theta}$

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