# Question 7318f

Jul 14, 2016

The question has not been written properly.

Given
$\csc x = 3 , \text{ where } {90}^{\circ} < x < {180}^{\circ}$

So ${45}^{\circ} < \frac{x}{2} < {90}^{\circ}$

$\implies \sin x = \frac{1}{3}$

${90}^{\circ} < x < {180}^{\circ} \to \cos x \text{ is negative}$

$\therefore \cos x = - \sqrt{1 - {\sin}^{2} x} = - \sqrt{1 - \frac{1}{9}}$
$= - \frac{2 \sqrt{2}}{3}$

$\text{Now } \cos \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 + \cos x\right)}$

$= \sqrt{\frac{1}{2} \left(1 - \frac{2 \sqrt{2}}{3}\right)}$

$= \sqrt{\frac{1}{6} \left(3 - 2 \sqrt{2}\right)}$

=sqrt(1/6((sqrt2)^2+1^2-2sqrt2*1)

=sqrt(1/6(sqrt2-1)^2

$= \frac{1}{\sqrt{6}} \left(\sqrt{2} - 1\right)$

$\text{Again } \sin \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 - \cos x\right)}$

$= \sqrt{\frac{1}{2} \left(1 + \frac{2 \sqrt{2}}{3}\right)}$

=sqrt(1/6(3+2sqrt2)

=sqrt(1/6((sqrt2)^2+1^2+2sqrt2*1)

=sqrt(1/6(sqrt2+1)^2#

$= \frac{1}{\sqrt{6}} \left(\sqrt{2} + 1\right)$

$\therefore \tan \left(\frac{x}{2}\right) = \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

$= \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$

$= {\left(\sqrt{2} + 1\right)}^{2} / \left(2 - 1\right) = 3 + 2 \sqrt{2}$