Question

$1+\cot \theta \le \cot \frac{\theta }{2}for0<\theta <\pi$. Find $\theta$ when equality sign holds.

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$1+\cot \theta =\cot \frac{\theta }{2}\Rightarrow 1+\frac{\cos \theta }{\sin \theta }=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}(\sin \frac{\theta }{2},\sin \theta \ne 0)$
$\Rightarrow \frac{\sin \theta +\cos \theta }{\sin \theta }=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\Rightarrow \sin \theta \sin \frac{\theta }{2}+\cos \theta \sin \frac{\theta }{2}=\cos \frac{\theta }{2}\sin \theta$
$\Rightarrow \sin \theta \sin \frac{\theta }{2}+\cos \theta \sin \frac{\theta }{2}-\cos \frac{\theta }{2}\sin \theta =0$

$\Rightarrow \sin \theta \sin \frac{\theta }{2}+\sin (\frac{\theta }{2}-\theta )=0\Rightarrow \sin \theta \sin \frac{\theta }{2}-\sin \frac{\theta }{2}=0$

$\Rightarrow \sin \frac{\theta }{2}\left(\sin \theta -1\right)=0\Rightarrow \sin \frac{\theta }{2}=0$ or $\sin \theta =1\Rightarrow \theta =\frac{\pi }{2}$