Q11:

Find the values of each of the expressions: $\tan^{-1}(\tan(\frac{3\pi}{4}))$

Step 1: Analyzing the Domain and Range Constraints

To evaluate $\tan^{-1}(\tan(x))$, we must recall that the property $\tan^{-1}(\tan(x)) = x$ holds if and only if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value branch of the tangent function [Per the definition of Inverse Trigonometric Functions].

In this problem, $x = \frac{3\pi}{4}$. Since $\frac{3\pi}{4} \approx 135^\circ$ and the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ corresponds to $(-90^\circ, 90^\circ)$, clearly $\frac{3\pi}{4} \notin (-\frac{\pi}{2}, \frac{\pi}{2})$.

Step 2: Utilizing Trigonometric Periodicity and Identity

Since the tangent function has a period of $\pi$, we use the identity $\tan(\pi - \theta) = -\tan(\theta)$. Furthermore, noting that $\tan(\theta) = \tan(\theta + n\pi)$, we look to simplify the expression using the supplementary angle property: $\tan(\pi - \theta) = -\tan(\theta)$. Alternatively, we use the specific identity:

$\tan(\frac{3\pi}{4}) = \tan(\pi - \frac{\pi}{4})$

Using the quadrant rule [Per the CAST rule for trigonometric functions in the second quadrant], we know that $\tan(\pi - \theta) = -\tan(\theta)$. Thus:

Step 3: Finding the Principal Value

We now substitute the value back into the inverse function:

By definition, $\tan^{-1}(-1) = y$ implies $\tan(y) = -1$, where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan(\frac{\pi}{4}) = 1$, and the tangent function is an odd function [Per the property $\tan(-x) = -\tan(x)$], we have:

Therefore, $y = -\frac{\pi}{4}$.

Visual Representation of the Transformation

Summary of Calculation

Final Solution: -\frac{\pi}{4}