Answer :
Solution
For this case we know that an angle theta is in the I quadrant and additionally:
[tex]\cos \theta=\frac{4}{\sqrt[]{23}}[/tex]We can find the sin of the angle usingt this identity:
[tex]\sin ^2\theta=1-\cos ^2\theta[/tex][tex]\sin \theta=\sqrt[]{1-(\frac{4}{\sqrt[]{23}})^2}=\sqrt[]{\frac{7}{23}}[/tex]We can find tangent:
[tex]\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt[]{\frac{7}{23}}}{\frac{4}{\sqrt[]{23}}}=\frac{\sqrt[]{7}}{4}[/tex]Finally we can find tan 2 theta and we have:
[tex]\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{2\cdot\frac{\sqrt[]{7}}{4}}{1-\frac{7}{16}}=\frac{\frac{\sqrt[]{7}}{2}}{\frac{9}{16}}=\frac{8\sqrt[]{7}}{9}[/tex]