Q.Find the value of $\tan^{-1}\left(\tan\frac{5\pi}{6}\right)+\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$.

Reduce each angle to its function's principal branch before evaluating.

Term 1: $\tan^{-1}$ has principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, and $\frac{5\pi}{6}$ lies outside it. Since $\tan$ has period $\pi$, $\tan\frac{5\pi}{6}=\tan\left(\frac{5\pi}{6}-\pi\right)=\tan\left(-\frac{\pi}{6}\right)$, and $-\frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. So $\tan^{-1}\left(\tan\frac{5\pi}{6}\right)=-\frac{\pi}{6}$.

Term 2: $\cos^{-1}$ has principal range $[0,\pi]$. Since $\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$, $\cos\frac{13\pi}{6}=\cos\frac{\pi}{6}$, and $\frac{\pi}{6}\in[0,\pi]$. So $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\frac{\pi}{6}$.

Add: $-\frac{\pi}{6}+\frac{\pi}{6}=0$.

The idea

$\tan^{-1}(\tan\theta)=\theta$ and $\cos^{-1}(\cos\theta)=\theta$ hold only when $\theta$ already sits in the function's principal range. When it does not, we replace $\theta$ by a period-shifted angle that has the same trig value but does lie in the principal range.

Term 1: $\tan^{-1}\left(\tan\frac{5\pi}{6}\right)$

The principal range of $\tan^{-1}$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, and $\frac{5\pi}{6}$ is outside it. Tangent has period $\pi$, so

Now $-\frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so

Term 2: $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$

The principal range of $\cos^{-1}$ is $[0,\pi]$. Cosine has period $2\pi$, and $\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$, so

Since $\frac{\pi}{6}\in[0,\pi]$,

Add

Method: Reducing $f^{-1}(f(\theta))$ to the principal branch

The general rule for any $\tan^{-1}(\tan\theta)$, $\cos^{-1}(\cos\theta)$, $\sin^{-1}(\sin\theta)$ term: the answer is not automatically $\theta$ — you must bring $\theta$ into the outer function's principal range while keeping the trig value fixed.

Steps

Step 1: State the principal range of the OUTER inverse function.

Step 2: Check whether the inside angle already lives there.

If $\theta$ is inside that range, the term is simply $\theta$ and you are done.

Step 3: If not, shift by a full period to an equivalent angle.

Use the period of the inner function — $\pi$ for tangent, $2\pi$ for sine and cosine — to replace $\theta$ by an angle with the same trig value that does lie in the principal range:

For cosine you may also need evenness, $\cos(-\alpha)=\cos\alpha$, to land in $[0,\pi]$.

Step 4: Read off the reduced angle and combine the terms.

Common Mistakes

Mistake 1: Writing $\tan^{-1}\!\big(\tan\tfrac{5\pi}{6}\big)=\tfrac{5\pi}{6}$.

Why it's wrong: $\tfrac{5\pi}{6}$ is outside the arctan range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, so the cancellation is invalid. Correct approach: subtract the period $\pi$ to get $\tfrac{5\pi}{6}-\pi=-\tfrac{\pi}{6}$, which is in range.

Mistake 2: Writing $\cos^{-1}\!\big(\cos\tfrac{13\pi}{6}\big)=\tfrac{13\pi}{6}$.

Why it's wrong: $\tfrac{13\pi}{6}$ exceeds $\pi$, so it is not in the arccos range $[0,\pi]$. Correct approach: subtract $2\pi$ first — $\tfrac{13\pi}{6}-2\pi=\tfrac{\pi}{6}$, which lies in $[0,\pi]$.

Mistake 3: Using the wrong period for the reduction.

Why it's wrong: tangent has period $\pi$ but sine and cosine have period $2\pi$; mixing them up gives an angle with the wrong value. Correct approach: shift $\tan$ arguments by $\pi$ and $\cos$ arguments by $2\pi$.