Question

The value of $\tan^{2}(\sec^{-1}(3))$ is

• A. 8
• B. 4
• C. 6
• D. 1
• E. 10

Hint:

Logic Tip: You can also solve this by sketching a right triangle. If $\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{3}{1}$, then by Pythagorean theorem, the opposite side is $\sqrt{3^2 - 1^2} = \sqrt{8}$. Thus, $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{8}}{1}$, and $\tan^2\theta = 8$.

Answer 1

The Correct Option is A

Concept:
This problem can be solved quickly by relating the trigonometric functions through the fundamental Pythagorean identity: $$\tan^2\theta = \sec^2\theta - 1$$
Step 1: Define a substitution variable.
Let $\theta = \sec^{-1}(3)$. By the definition of an inverse function, this implies: $$\sec\theta = 3$$
Step 2: Substitute $\theta$ into the target expression.
We are tasked with evaluating $\tan^2(\sec^{-1}(3))$. Using our substitution, this becomes: $$\tan^2\theta$$
Step 3: Apply the Pythagorean identity.
Convert the expression into terms of secant: $$\tan^2\theta = \sec^2\theta - 1$$ Substitute the known value of $\sec\theta = 3$: $$\tan^2\theta = (3)^2 - 1$$ $$\tan^2\theta = 9 - 1 = 8$$