Step 1: Evaluate \( \tan^2(\sec^{-1}4) \) Let \( \theta = \sec^{-1}4 \). This implies \( \sec(\theta) = 4 \). Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we have \( \cos(\theta) = \frac{1}{4} \). Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we get \( \sin^2(\theta) = 1 - \left( \frac{1}{4} \right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \). Thus, \( \sin(\theta) = \frac{\sqrt{15}}{4} \). Then, \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}} = \sqrt{15} \). Therefore, \( \tan^2(\theta) = (\sqrt{15})^2 = 15 \).Step 2: Evaluate \( \cot(\csc^{-1}3) \) Let \( \theta = \csc^{-1}3 \). This implies \( \csc(\theta) = 3 \). Since \( \csc(\theta) = \frac{1}{\sin(\theta)} \), we have \( \sin(\theta) = \frac{1}{3} \). Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we get \( \cos^2(\theta) = 1 - \left( \frac{1}{3} \right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \). Thus, \( \cos(\theta) = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3} \). Then, \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = 2\sqrt{2} \).Step 3: Sum the results Adding the results from Step 1 and Step 2: \( 15 + 2\sqrt{2} \).Answer: The value of the expression is \( 15 + 2\sqrt{2} \). Therefore, the correct answer is option (2).