Question

lf $\tan \frac{\theta }{2}=cosec\theta -\sin \theta$, then the numerical value of ${\cos }^2\frac{\theta }{2}$ is

• A. $\frac{\sqrt{3}-1}{4}$
• B. $\frac{\sqrt{5}+1}{4}$
• C. $\frac{\sqrt{3}+1}{4}$
• D. $\frac{\sqrt{5}-1}{4}$

Answer 1

The correct option is B $\frac{\sqrt{5}+1}{4}$

we get straight away
$\tan \frac{\Theta }{2}=\frac{{\cos }^2\Theta }{\sin \Theta }$
so $\left(\tan \frac{\Theta }{2}\right)\left(2\sin \frac{\Theta }{2}\cos \frac{\Theta }{2}\right)={\cos }^2\Theta$
so $2{\sin }^2\frac{\Theta }{2}={\cos }^2\Theta$
so $2-2{\cos }^2\frac{\Theta }{2}=(2{\cos }^2\frac{\Theta }{2}-1{)}^2$
we solve for ${\cos }^2\Theta$