Find an algebraic expression equivalent to $\sec \left(\tan ^{-1} x\right)$
A) $\sqrt{x^{2}+1}$
B) $\frac{1}{\sqrt{x^{2}+1}}$
C) $\frac{\sqrt{x^{2}+1}}{x}$
D) $\sqrt{x^{2}-1}$
E) $x^{2}+1$

Question

Quizplus · Accepted Answer

To find an expression equivalent to $\sec(\tan^{-1} x)$, we start by considering a right triangle where the angle $\theta$ satisfies $\tan(\theta) = x$. In this triangle, the opposite side to $\theta$ is $x$ and the adjacent side is $1$, making the hypotenuse $\sqrt{x^2 + 1}$ by the Pythagorean theorem. Since $\sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}}$, $\sec(\tan^{-1} x) = \frac{\sqrt{x^2 + 1}}{1} = \sqrt{x^2 + 1}$.

Find an Algebraic Expression Equivalent To sec⁡(tan⁡−1x)\sec \left(\tan ^{-1} x\right)sec(tan−1x) A) x2+1\sqrt{x^{2}+1}x2+1​

Find an Algebraic Expression Equivalent To $\sec \left(\tan ^{-1} x\right)$ A) $\sqrt{x^{2}+1}$