Evaluating $\sec(\tan^{-1}\sqrt3)$ exactly

Question

Find the exact value of the expression $$\sec(\tan^{-1}\sqrt3)$$

I'm confused about how to start the problem and what quadrant to use.

Answer 1

By definition, $\tan^{-1}\sqrt(3)$ is in quad $1$ ( since $\sqrt(3) > 0$), so $\sec(\tan^{-1}\sqrt(3))=\frac{2}{1}~(>0)$.

Answer 2

HINT

\begin{align*} \sec^{2}(x) = \tan^{2}(x) + 1 \end{align*}

Answer 3

The best way to start is to recall the definition: $$ \sec x=\frac1{\cos x}. $$ Since the range of $\tan^{-1}(x)$ is $[-\frac\pi2,\frac\pi2]$ we conclude that the value of cosine is positive. Hence: $$ \frac1{\cos x}=\sqrt{1+\tan^2x}=2. $$