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Math Help - Trigonometric identity issue!

  1. #1
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    Trigonometric identity issue!

    Hi, currently questioning the validity of my method solving an identity, not sure if this is the correct way or not?
    Firstly I squared both sides, then proceeded to multiply through by cos^2(A/2), to then obtain the left hand side.
    Is this ok and is there a faster/more efficient method? Would appreciate greatly any feedback.Thanks.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric identity issue!

    My approach would be to begin with the left side, and work to get the right:

    \sqrt{\frac{1-\sin(A)}{1+\sin(A)}}\cdot\sqrt{\frac{1+\sin(A)}{1+  \sin(A)}}=\frac{\cos(A)}{1+\sin(A)}=

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

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

    \frac{1-\tan\left(\frac{A}{2} \right)}{1+\tan\left(\frac{A}{2} \right)}
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  3. #3
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    Re: Trigonometric identity issue!

    Trigonometric identity issue!-trig.png
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