Thread: How do I get the final solution?

Hi

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

Originally Posted by Sunyata

Hi

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

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

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

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

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

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

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

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