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Math Help - difficult question!!!!

  1. #1
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    difficult question!!!!

    \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
    \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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    \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}=

    =\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)}=

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

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

    Then the expression is 0.
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