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Math Help - trig solvein problem

  1. #1
    Member srirahulan's Avatar
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    trig solvein problem

    \sqrt\frac{1+sinA}{1-sinA}=tan[\frac{\pi}{4}+\frac{A}{2}] when i simplify finally i get \frac{1+sinA}{cosa}then how can i prove it with square root.
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  2. #2
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    Re: trig solvein problem

    \tan\left(\frac{\pi}{4} + \frac{A}{2}\right)

    \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{A}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right) \cdot  \tan\left(\frac{A}{2}\right)}

    \frac{1+\tan\left(\frac{A}{2}\right)}{1-\tan\left(\frac{A}{2}\right)}

    \frac{\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2  }\right) - \sin\left(\frac{A}{2}\right)} \cdot \frac{\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2  }\right) + \sin\left(\frac{A}{2}\right)}

    \frac{1 + 2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}  \right)}{\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)}

    \frac{1 + \sin{A}}{\cos{A}}
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  3. #3
    Member srirahulan's Avatar
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    Re: trig solvein problem

    i simplify this by the r.h.s and i finally get (1+sina)/cosa then how can i show this as in the l.h.s (in square root)
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  4. #4
    MHF Contributor
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    Re: trig solvein problem

    work backwards ... you figure it out for once.
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