Question

If tan theta = 2square root of 14/5 and pi

Answer 1

Given:

$\tan (\theta)=\frac{2\sqrt[]{14}}{5},\pi<\theta<\frac{3\pi}{2}$

Use the identity,

$\begin{gathered} \tan \theta=\frac{2\tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})} \\ \text{Let }\tan (\frac{\theta}{2})=x \\ \tan \theta=\frac{2x}{1-x^2} \end{gathered}$

Simplify,

$\begin{gathered} \tan \theta=\frac{2x}{1-x^2} \\ (1-x^2)\tan \theta=2x \\ -\tan \theta x^2+\tan \theta=2x \\ \tan \theta x^2+2x-\tan \theta=0 \end{gathered}$

Use the quadratic formula,

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=\tan \theta,b=2,c=-tan\theta \\ x=\frac{-2\pm\sqrt[]{2^2-4(\tan\theta)(-\tan\theta)}}{2(\tan\theta)} \\ x=\frac{-2\pm2\sqrt[]{1+\tan ^2\theta}}{2\tan \theta} \\ x=\frac{-1\pm\sqrt[]{1+\tan^2\theta}}{\tan \theta} \end{gathered}$$\begin{gathered} \text{Given that: }\tan (\theta)=\frac{2\sqrt[]{14}}{5} \\ x=\frac{-1\pm\sqrt[]{1+(\frac{2\sqrt[]{14}}{5})^2}}{\frac{2\sqrt[]{14}}{5}} \\ x=\frac{-1\pm\sqrt[]{1+\frac{4\cdot14}{25}^{}}}{\frac{2\sqrt[]{14}}{5}} \\ x=\frac{-1+\frac{9}{5}^{}}{\frac{2\sqrt[]{14}}{5}},x=\frac{-1-\frac{9}{5}^{}}{\frac{2\sqrt[]{14}}{5}} \\ x=\frac{4}{2\sqrt[]{14}},x=\frac{-14}{2\sqrt[]{14}} \\ x=\frac{2}{\sqrt[]{14}}.x=-\frac{7}{\sqrt[]{14}} \end{gathered}$

It gives,

$\begin{gathered} \tan (\frac{\theta}{2})=\frac{2}{\sqrt[]{14}},\tan (\frac{\theta}{2})=-\frac{7}{\sqrt[]{14}} \\ \text{Also,}\pi<\theta<\frac{3\pi}{2}\text{ , lies in 3rd quadrant} \\ \tan \theta\text{ is positive in 3rd quadrant } \\ \Rightarrow\tan (\frac{\theta}{2})=\frac{2}{\sqrt[]{14}} \end{gathered}$

Answer:

$\tan (\frac{\theta}{2})=\frac{2}{\sqrt[]{14}},\pi<\theta<\frac{3\pi}{2}$