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Thread: difficult question!!!!

  1. #1
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    difficult question!!!!

    $\displaystyle \log_{\tan(\frac{\pi}{4}+\frac{x}{2})} (\sec x + \tan x)-\log_{\tan(\frac{\pi}{4}-\frac{x}{2})} (\sec x - \tan x)$
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  2. #2
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    More information please

    Hello jashansinghal
    Quote Originally Posted by jashansinghal View Post
    $\displaystyle \log_{\tan(\frac{\pi}{4}+\frac{x}{2})} (\sec x + \tan x)-\log_{\tan(\frac{\pi}{4}-\frac{x}{2})} (\sec x - \tan x)$
    So what is the question?

    Grandad
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  3. #3
    MHF Contributor red_dog's Avatar
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    $\displaystyle \sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}=\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin \frac{x}{2}\cos\frac{x}{2}}{\cos x}=$

    $\displaystyle =\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right )^2}{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\f rac{x}{2}\right)}=$

    $\displaystyle =\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{ x}{2}-\sin\frac{x}{2}}=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\tan\left(\frac{\pi}{4}+\frac{x}{ 2}\right)$

    Then $\displaystyle \log_{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}( \sec x+\tan x)=1$

    In the same way, $\displaystyle \log_{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}(\sec x-\tan x)=1$

    Then the expression is 0.
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