Question

Prove that $\frac{a\sin \theta }{a(1+\sin \theta )}=\frac{2\tan \frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}+2\tan \frac{\theta }{2}}$

Answer 1

$\frac{a\sin \theta }{a(1+\sin \theta )}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{({\sin }^2\frac{\theta }{2}+{\cos }^2\frac{\theta }{2}+2\sin \frac{\theta }{2}\cos \frac{\theta }{2})}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{{(\sin \frac{\theta }{2}+\cos \frac{\theta }{2})}^2}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}{(\tan \frac{\theta }{2}+1)}^2}$

$=\frac{2\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}{(\tan \frac{\theta }{2}+1)}^2}$

$=\frac{2\tan \frac{\theta }{2}}{{(\tan \frac{\theta }{2}+1)}^2}$

$=\frac{2\tan \frac{\theta }{2}}{{\tan }^2\frac{\theta }{2}+1+2\tan \frac{\theta }{2}}$

$=\frac{2\tan \frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}+2\tan \frac{\theta }{2}}$