Q.Find the principal value of the following: $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$

Since $\tan^{-1}$ must return an angle from its principal value branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\frac{3\pi}{4}$ lies outside that interval, we cannot simply cancel the two operations — we first evaluate the inside, then bring the result back into the branch.

Step 1: $\tan \frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan \frac{\pi}{4} = -1$. …

The inverse tangent function, $\tan^{-1}x$ (also written as $\arctan x$), is defined to return a principal value — a single, unambiguous angle. For $\tan^{-1}$, the agreed-upon range is $(-\pi/2, \pi/2)$, i.e., strictly between $-90^\circ$ and $90^\circ$. This is the interval where the tangent function is one-to-one, so we can invert it cleanly.

The expression $\tan^{-1}(\tan \theta)$ looks like it should just give back $\theta$, but that's only true if $\theta$ itself already lies in $(-\pi/2, \pi/2)$. If $\theta$ is outside that interval, the function first evaluates $\tan \theta$, then asks: "What angle inside $(-\pi/2, \pi/2)$ has this same tangent value?" That's the principal value.

Here, $\theta = \frac{3\pi}{4}$. Let's see where that sits.

1. Locate the angle. $\frac{3\pi}{4}$ is $135^\circ$, which lies in the second quadrant. Its tangent is negative there (since tangent = sine/cosine, and in QII sine is positive, cosine is negative). Specifically:

2. Now find the principal value.

   We need an angle $\alpha$ in $(-\pi/2, \pi/2)$ such that $\tan \alpha = -1$.

   The angle whose tangent is $-1$ in that interval is $-\pi/4$ (i.e., $-45^\circ$).

   Check: $\tan(-\pi/4) = -1$, and $-\pi/4 \in (-\pi/2, \pi/2)$.

3. Why not $3\pi/4$ itself? …

Method: Evaluating $\tan^{-1}(\tan\theta)$

Use this to simplify $\tan^{-1}(\tan\theta)$ when $\theta$ might fall outside the principal range of $\tan^{-1}$.

Steps

Step 1: Recall the principal range.

$\tan^{-1}$ returns an angle strictly inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. If $\theta$ is already there, the answer is $\theta$.

Step 2: Exploit the period of tangent, which is $\pi$ (not $2\pi$).

Since $\tan(\theta - k\pi) = \tan\theta$ for any integer $k$, subtract or add whole multiples of $\pi$ to slide $\theta$ into $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: …

Common Mistakes

Mistake 1: Writing $\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \frac{3\pi}{4}$.

Why it's wrong: $\frac{3\pi}{4}$ lies outside the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so it cannot be the output. Correct approach: subtract $\pi$ to reach $-\frac{\pi}{4}$, which is in range and has the same tangent.

Mistake 2: Getting the sign wrong and answering $+\frac{\pi}{4}$. …