Solution

I believe one of the most critical to remember aspects of inverse trig functions is to remember that the inverse trig function always returns an angle: \(\theta = \tan^{-1}\left(-\frac{ 7 }{ 2 }\right)\). This also allows us to naturally consider the tangent version of this equation: \(\tan(\theta)=-\frac{ 7 }{ 2 }\). In doing so, we gain some insight into how to draw a triangle which represents the angle \(\theta\).

To determine how/where to draw this triangle, first let's consider what the Inverse Tangent's Domain/Range are. On the one hand, it has a domain of all real numbers, while on the other its range is \( \left(-\frac{\pi}{{2}},\frac{\pi}{{2}}\right)\). Since the input is \(-\frac{ 7 }{ 2 }\), we know that we need the tangent to be negative, which on the Unit Circle puts us in either quadrant II or quadrant IV. However, the range restriction forces us to be in quadrant 4 since we cannot have angles greater that \(\pi/2\) (which is what we would need to land in the 2nd quadrant). Once we have determined the appropriate quadrant, we can then sketch a generic right triangle. Since we are looking at the tangent inverse function, the easiest triangle to draw is one where we treat the \(-7/2\) as literally the Opposite over Adjacent of a right triangle, in quadrant IV:

All that remains is to solve the Hypotenuse using the Pythagorean Theorem, and then apply the Secant to the angle \(\theta\): \[\solve{ H^2 &=& 2^2 +(-7)^2\\ H^2 &=& 4 + 49\\ H^2 &=& 53\\ H &=& \sqrt{{53}} } \]Since the Hypotenuse is the length of the triangle, we take only the positive square root. The Secant, being the reciprocal of Cosine, can be determined from a right triangle as the Hypotenuse over the Adjacent side: \[ \sec\left(\tan^{-1}\left(-\frac{ 7 }{ 2 }\right)\right) = \sec(\theta) = \frac{\sqrt{ 53 } }{ 2 } \]