Trigonometric identity issue!

**DonGorgon** · February 18th 2013, 01:41 PM

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.

**MarkFL** · February 18th 2013, 03:44 PM

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

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

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

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

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

**ibdutt** · February 18th 2013, 07:44 PM

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