Question

$\frac{1}{1-\cos \theta +2i\sin \theta }=\frac{1-2i\cot \left(\theta /2\right)}{5+3\cos \theta }$. If this is true enter 1, else enter 0.

Answer 1

LHS is

$\frac{1}{1-\cos \theta +2i\sin \theta }=\frac{1-\cos \theta -2i\sin \theta }{(1-\cos \theta {)}^2+4{\sin }^2\theta }$

$=\frac{2{\sin }^2\frac{\theta }{2}-4i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1+{\cos }^2\theta -2\cos \theta +4{\sin }^2\theta }$ [As $1-\cos 2A=2{\sin }^{2}A$ and $\sin 2A=2\sin A\cos A$]

$=\frac{2{\sin }^2\frac{\theta }{2}-4i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2-2\cos \theta +3{\sin }^2\theta }$ (Using ${\sin }^2\theta +{\cos }^2\theta =1$)

$=\frac{2{\sin }^2\frac{\theta }{2}-4i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{4{\sin }^2\frac{\theta }{2}+3{\sin }^2\theta }$ $=\frac{2{\sin }^2\frac{\theta }{2}-4i\sin \frac{\theta }{2}cos\frac{\theta }{2}}{4{\sin }^2\frac{\theta }{2}+3\left(4{\sin }^2\frac{\theta }{2}{\cos }^2\frac{\theta }{2}\right)}$

Divide both numerator and dinominator by ${\sin }^2\frac{\theta }{2}$

$\frac{2-4i\cot \frac{\theta }{2}}{4+12{\cos }^2\frac{\theta }{2}}=\frac{1-2i\cot \frac{\theta }{2}}{2+6{\cos }^2\frac{\theta }{2}}=\frac{1-2i\cot \frac{\theta }{2}}{2+3(1+\cos \theta )}=\frac{1-2i\cot \frac{\theta }{2}}{5+3\cos \theta }\to$ $RHS$

$\therefore$ $LHS=RHS$

$\therefore$ The answer is $1$.