Question

The general solution of the equation $2\cot \frac{\theta }{2}={(1+\cot \theta )}^2$ is

• A. $n\pi +{(-1)}^n\frac{\pi }{4},nϵz$
• B. $n\pi +{(-1)}^n\frac{\pi }{3},nϵz$
• C. $n\pi +{(-1)}^n\frac{\pi }{6},nϵz$
• D. None

Answer 1

Answer: (C)

$n\pi +{(-1)}^n\frac{\pi }{6},nϵz$

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

$\Rightarrow 2{\sin }^2\theta -\sin \theta =0$

$\Rightarrow \sin \theta (2\sin \theta -1)=0$

$\Rightarrow \sin \theta =0or\sin \theta =\frac{1}{2}$

The general solution are $\theta =m\pi ,m\in z$ , $\theta =n\pi +{(-1)}^n\frac{\pi }{6},n\in z$