Question

If $\tan \frac{\theta }{2}=cosec\theta -\sin \theta$, then

• A. ${\sin }^2\frac{\theta }{2}=2{\sin }^2{18}^o$
• B. $\cos 2\theta +2\cos \theta +1=0$
• C. ${\sin }^2\frac{\theta }{2}=4{\sin }^2{18}^o$
• D. $\cos 2\theta +2\cos \theta -1=0$

Answer 1

Answer: (A), (D)

The correct options are
A ${\sin }^2\frac{\theta }{2}=2{\sin }^2{18}^o$
D $\cos 2\theta +2\cos \theta -1=0$

Simplifying, we get
$\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{1-{\sin }^2\theta }{\sin \theta }$

$\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{{\cos }^2\left(\theta \right)}{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}}$

$2{\sin }^2\frac{\theta }{2}={\cos }^2\theta$

$2{\sin }^2\frac{\theta }{2}=(1-2{\sin }^2\frac{\theta }{2}{)}^2$
Let, $2{\sin }^2\frac{\theta }{2}=t$
$⟹t=(1-t{)}^2$
$⟹t=t^2-2t+1$
$⟹t^2-3t+1=0$
$⟹t=\frac{3\pm \sqrt{9-4}}{2}$

$⟹t=\frac{3\pm \sqrt{5}}{2}$
Now $|\sin \theta |\le 1$
$⟹t=\frac{3-\sqrt{5}}{2}$
Or
${\sin }^2\frac{\theta }{2}=\frac{3-\sqrt{5}}{4}$
$⟹2\left(\frac{3-\sqrt{5}}{8}\right)=2{\sin }^2\left({18}^0\right)$
Hence, option A is correct.
${\cos }^2\frac{\theta }{2}=1-{\sin }^2\frac{\theta }{2}=\frac{1+\sqrt{5}}{4}$
Option D
$\cos 2\theta +2\cos \theta -1=0$
$-2{\sin }^2\theta +2\cos \theta =0$
$=-2+2{\cos }^2\theta +2\cos \theta =0$
$=-1+{\cos }^2\theta +\cos \theta =0$
${\cos }^2\theta =1-\cos \theta =2{\sin }^2\frac{\theta }{2}$ which is true (as proven in the first part)

Hence, option A and D are correct.