Question

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

Answer 1

Given:

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

Use the identity,

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

Simplify,

`\begin{gathered} \tan \theta=(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+(4\cdot14)/(25)^{}}}{\frac{2\sqrt[]{14}}{5}} \\ x=\frac{-1+(9)/(5)^{}}{\frac{2\sqrt[]{14}}{5}},x=\frac{-1-(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 ((\theta)/(2))=\frac{2}{\sqrt[]{14}},\tan ((\theta)/(2))=-\frac{7}{\sqrt[]{14}} \\ \text{Also,}\pi<\theta<(3\pi)/(2)\text{ , lies in 3rd quadrant} \\ \tan \theta\text{ is positive in 3rd quadrant } \\ \Rightarrow\tan ((\theta)/(2))=\frac{2}{\sqrt[]{14}} \end{gathered}`

Answer:

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