Question

Write the expression in algebraic form. \(\tan \left(\operatorname{arcsec} \frac{x}{3}\right)\)

Short answer

The desired expression, in algebraic form, is \(\frac{\sqrt{x^2 - 9}}{3}\)

Step-by-step solution

Expressing \(\operatorname{arcsec}\) as a right-angle triangle

Let's name an angle \(Θ = \operatorname{arcsec} \frac{x}{3}\). In terms of a right-angle triangle, \(\operatorname{sec}(Θ) = \frac{x}{3}\). The secant of an angle in a right-angle triangle is the ratio of the hypotenuse to the adjacent side. Therefore, we can choose the hypotenuse to be 'x' and the adjacent side to be '3', with the remaining side (opposite side) unknown.

Calculating the length of the remaining side

By using the Pythagorean theorem \(c^2 = a^2 + b^2\), where 'c' is the hypotenuse, and 'a' and 'b' are the other two sides of the triangle, we can calculate the length of the remaining (opposite) side, let's call it 'y'. So \(x^2 = 3^2 + y^2\), this gives \(y^2 = x^2 - 9\). Thus, \(y = \sqrt{x^2 - 9}\). However, it isn't necessary to take the negative root as 'x' is larger than '3'

Express the original quantity in terms of the triangle

The tanget of an angle in a right-angle triangle is the ratio of the opposite side to the adjacent side. So, we can express the initial given function as \(\tan(Θ) = \frac{y}{3} = \frac{\sqrt{x^2 - 9}}{3}\).