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sin⁡(a)cos⁡(a)cos⁡(a)(sin⁡(a)−sin⁡(a)cos⁡(a))=sin⁡(a)1−cos⁡(a)\frac{\sin\left(a\right)\cos\left(a\right)}{\cos\left(a\right)\left(\sin\left(a\right)-\sin\left(a\right)\cos\left(a\right)\right)}=\frac{\sin\left(a\right)}{1-\cos\left(a\right)}

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Final answer to the exercise

a=0+2πn, a=2π+2πn, a=12π+2πn ,  n∈Za=0+2\pi n,\:a=2\pi+2\pi n,\:a=\frac{1}{2}\pi+2\pi n\:,\:\:n\in\Z

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sin(a)cos(a)cos(a)(sin(a)−sin(a)cos(a)) =sin(a)1−cos(a) 
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