​$$\text{If } \sec \theta = \frac{4}{3}, \text{ what is the value of } \tan^2 \theta + \tan^4 \theta \text{?}$$​​

Correct option is A

​$$\textbf{\underline{Given:} } \\\sec \theta = \frac{4}{3}\\\\\textbf{\underline{Concept Used: }} \\ \sec^2 \theta = 1 + \tan^2 \theta\\\ \\\textbf{\underline {Solution: }} \\ \sec^2 \theta = \left(\frac{4}{3}\right)^2 = \frac{16}{9}\\ \\\Rightarrow \frac{16}{9} = 1 + \tan^2 \theta\\ \\\Rightarrow \tan^2 \theta = \frac{16}{9} - 1\\\ \\\Rightarrow \tan^2 \theta = \frac{16}{9} - \frac{9}{9}\\\ \\\Rightarrow \tan^2 \theta = \frac{7}{9}\\\ \\\text{Now, calculate } \tan^4 \theta:\\\ \\\tan^4 \theta = \left(\tan^2 \theta\right)^2 = \left(\frac{7}{9}\right)^2 = \frac{49}{81}\\\ \\\text{Now, find } \tan^2 \theta + \tan^4 \theta:\\\ \\\tan^2 \theta + \tan^4 \theta = \frac{7}{9} + \frac{49}{81}\\\ \\\Rightarrow \tan^2 \theta + \tan^4 \theta = \frac{63}{81} + \frac{49}{81} = \frac{112}{81}\\\ \\\text{Therefore, the value is } \boxed{\frac{112}{81}}.\\$$​​