Discussion
The value of \(tan^{2}(sec^{-1}(3))\) is
Answer: A
The value is 8.
Let \(\theta = \sec^{-1}3\), so \(\sec\theta = 3\). Then use \(\tan^{2}\theta = \sec^{2}\theta - 1\).
\[ \tan^{2}\theta = 3^{2} - 1 = 8 \]
Using \(\sec\theta - 1\) without squaring gives 2, and adding instead of subtracting gives 10, the trap in the options. Drawing a right triangle with hypotenuse 3 and adjacent side 1 gives opposite side \(\sqrt{8}\), so \(\tan\theta = \sqrt{8}\) and its square is 8.
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