Solution
We are told that
[tex]\theta\text{ is in the first quadrant}[/tex]and the standard position with P(u,v) = (3,4)
We want to find
[tex]\tan 2\theta[/tex]First, We will draw the standard position
Notice that from the diagram above, the (3,4) indicates that the horizontal distance is 3 units and the vertical distance is 4 units
From the diagram above
Using SOHCAHTOA
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan \theta=\frac{4}{3} \end{gathered}[/tex]We are left with computing
[tex]\tan 2\theta[/tex]The addition formula for tan(A+B)
[tex]\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \text{Let A = B = }\theta \\ \tan (\theta+\theta)=\frac{\tan\theta+\tan\theta}{1-\tan^2\theta} \\ \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta} \end{gathered}[/tex]We now imput the values of tan
[tex]\begin{gathered} \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta} \\ \tan 2\theta=\frac{2(\frac{4}{3})}{1-(\frac{4}{3})^2} \\ \tan 2\theta=\frac{\frac{8}{3}}{1-\frac{16}{9}} \\ \tan 2\theta=\frac{\frac{8}{3}}{\frac{9}{9}-\frac{16}{9}} \\ \tan 2\theta=\frac{\frac{8}{3}}{-\frac{7}{9}} \\ \tan 2\theta=\frac{8}{3}\times-\frac{9}{7} \\ \tan 2\theta=-\frac{24}{7} \end{gathered}[/tex]Therefore, the correct answer is -24/7
Option C
