How do I get the final solution?

**Sunyata** · March 23rd 2010, 03:13 AM

Hi

Could someone please show me how to get from the LHS to the RHS???

**Archie Meade** · March 23rd 2010, 04:28 AM

Originally Posted by Sunyata

Hi

Could someone please show me how to get from the LHS to the RHS???

$\frac{atan\theta}{tan\theta-sec\theta+1}=\frac{a\frac{sin\theta}{cos\theta}}{\ frac{sin\theta}{cos\theta}-\frac{1}{cos\theta}+1}$

$=\frac{asin\theta}{sin\theta-1+cos\theta}=\frac{asin\left(\frac{2\theta}{2}\rig ht)}{sin\left(\frac{2\theta}{2}\right)+cos\left(\f rac{2\theta}{2}\right)-1}$

$=\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2 sin\frac{\theta}{2}cos\frac{\theta}{2}+cos^2\left( \frac{\theta}{2}\right)-sin^2\left(\frac{\theta}{2}\right)-1}$

$=\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2 sin\frac{\theta}{2}cos\frac{\theta}{2}+\left(1-sin^2\left(\frac{\theta}{2}\right)\right)-sin^2\left(\frac{\theta}{2}\right)-1}$

$=\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2 sin\frac{\theta}{2}cos\frac{\theta}{2}-2sin^2\left(\frac{\theta}{2}\right)}$

$=\frac{2acos\frac{\theta}{2}}{2cos\frac{\theta}{2}-2sin\frac{\theta}{2}}$

$=\frac{acos\frac{\theta}{2}}{cos\frac{\theta}{2}-sin\frac{\theta}{2}}$

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