If π < θ < (3π)/(2) and cos θ = - (3)/(5) , then ((θ)/(4))& is...

cos \theta = - \dfrac{3}{5}⇒ 2cos^{2}\dfrac{\theta}{2} - 1 = - \dfrac{3}{5} 2cos^{2}\dfrac{\theta}{2} = 1 - \dfrac{3}{5} = \dfrac{2}{5} cos\dfrac{\theta}{2} = \pm \dfrac{1}{\sqrt{5}} ⇒ cos\dfrac{\theta}{2} = - \dfrac{1}{\sqrt{5}} \left(∵ \dfrac{\pi}{2}$<$\dfrac{\theta}{2}$<$\dfrac{3\pi}{4}\right) Now, $tan^{2}$\dfrac{\theta}{4} = \dfrac{1 - cos\dfrac{\theta}{2}}{1 + cos\dfrac{\theta}{2}} = \dfrac{1 + \dfrac{1}{\sqrt{5}}}{1 - \dfrac{1}{\sqrt{5}}} = \dfrac{\sqrt{5} + 1}{\sqrt{5} - 1} $tan^{2}$\dfrac{\theta}{4} = \dfrac{{\left(\sqrt{5} + 1\right)}^{2}}{4} = {\left(\dfrac{\left(\sqrt{5} + 1\right)}{2}\right)}^{2} $⇒ tan$\dfrac{\theta}{4} = \dfrac{\sqrt{5} + 1}{2} \left(∵ \dfrac{\pi}{4}$<$\dfrac{\theta}{4}$<$\dfrac{3\pi}{8}\right)