Q.Find the value of $\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$.

Concept: Inverse Trigonometric Graphs – The inverse tangent function $\tan^{-1}$ returns the principal value in $(-\pi/2, \pi/2)$. So we must adjust the given angle to lie within this range.

1. Compute the angle: $\dfrac{7\pi}{6} = \pi + \dfrac{\pi}{6}$, which lies in the third quadrant.
2. The tangent function has period $\pi$, so $\tan\left(\dfrac{7\pi}{6}\right) = \tan\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$.
3. Now $\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$, since $\dfrac{\pi}{6}$ is the unique angle in $(-\pi/2, \pi/2)$ with that tangent.

1. The core idea: Inverse functions need the right domain

The function $\tan^{-1}(\tan x)$ is not simply $x$ for every $x$. Why? Because $\tan x$ is periodic with period $\pi$, so many different $x$ values give the same tangent. The inverse tangent, $\tan^{-1}$, is defined to return a unique angle — the one lying in the principal branch $(-\pi/2, \pi/2)$.

So when you see $\tan^{-1}(\tan x)$, the answer is the unique angle in $(-\pi/2, \pi/2)$ that has the same tangent as $x$.

2. Check where $\frac{7\pi}{6}$ lies

$\frac{7\pi}{6}$ is $210^\circ$. That's in the third quadrant, well outside $(-\pi/2, \pi/2)$ (which is $-90^\circ$ to $90^\circ$). So we cannot just cancel.

3. Find an equivalent angle in the principal branch

Since $\tan$ has period $\pi$, we can subtract $\pi$ from $\frac{7\pi}{6}$:

Now $\frac{\pi}{6}$ is $30^\circ$, which lies nicely inside $(-\pi/2, \pi/2)$. And crucially:

because adding or subtracting $\pi$ doesn't change the tangent value.

4. Apply the inverse

Now we have:

Since $\frac{\pi}{6}$ is in $(-\pi/2, \pi/2)$, the function $\tan^{-1}(\tan x)$ simply returns $x$ itself. So:

Method: Evaluating $\tan^{-1}(\tan\theta)$ using the period $\pi$

Use this to bring $\tan^{-1}(\tan\theta)$ into the principal range when $\theta$ is outside it.

Steps

Step 1: State the principal range.

$\tan^{-1}$ returns a value in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. If $\theta$ is already inside, stop — the answer is $\theta$.

Step 2: Shift by multiples of $\pi$, not $2\pi$.

Tangent repeats every $\pi$, so choose the integer $k$ with $\theta - k\pi \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$:

Step 3: Sanity-check the sign by quadrant.

Confirm the sign of the answer matches the sign of $\tan\theta$ (positive in Q1/Q3, negative in Q2/Q4).

Common Mistakes

Mistake 1: Subtracting $2\pi$ instead of $\pi$.

Why it's wrong: tangent has period $\pi$, not $2\pi$; subtracting $2\pi$ from $\frac{7\pi}{6}$ gives $-\frac{5\pi}{6}$, still outside the principal range. Correct approach: subtract $\pi$ to get $\frac{\pi}{6}$, which is in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Mistake 2: Answering $\frac{7\pi}{6}$ by cancelling directly.

Why it's wrong: $\frac{7\pi}{6}$ is outside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so it cannot be the principal value. Correct approach: since $\tan\frac{7\pi}{6} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, the answer is $\frac{\pi}{6}$.