Q.Find the value of $\tan^{-1}\left(\frac{-1}{\sqrt3}\right)+\cot^{-1}\left(\frac{1}{\sqrt3}\right)+\tan^{-1}\left[\sin\left(\frac{-\pi}{2}\right)\right]$.

Evaluate each inverse-trig term on its principal branch, then add.

Term 1: $\tan^{-1}\left(-\frac{1}{\sqrt3}\right)$. Since $\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt3}$ and $-\frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, this is $-\frac{\pi}{6}$.

Term 2: $\cot^{-1}\left(\frac{1}{\sqrt3}\right)$. Since $\cot\frac{\pi}{3}=\frac{1}{\sqrt3}$ and $\frac{\pi}{3}\in(0,\pi)$, this is $\frac{\pi}{3}$. …

The idea

Every inverse-trig function returns the unique angle in a fixed principal range. Read each term off that range, then combine.

Term 1: $\tan^{-1}\left(-\frac{1}{\sqrt3}\right)$

Principal range of $\tan^{-1}$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Because $\tan\frac{\pi}{6}=\frac{1}{\sqrt3}$ and tangent is odd, $\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt3}$. Hence

Term 2: $\cot^{-1}\left(\frac{1}{\sqrt3}\right)$

Principal range of $\cot^{-1}$ is $(0,\pi)$. Since $\cot\frac{\pi}{3}=\frac{\cos(\pi/3)}{\sin(\pi/3)}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}$ and $\frac{\pi}{3}\in(0,\pi)$,

$\cot^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{3}.$ …

Method: Evaluating a sum of several inverse-trig terms

When an expression adds up several inverse-trig values, evaluate each term separately on its own principal branch, then combine — never assume the terms cancel or telescope before you have their exact values.

Steps

Step 1: Simplify any ordinary trig nested inside an inverse first.

E.g. reduce $\sin\!\big(-\tfrac{\pi}{2}\big)=-1$ before taking $\tan^{-1}$ of it, so the inverse acts on a clean number.

Step 2: Evaluate each inverse term, stating its range.

$\tan^{-1}:\ \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right),\qquad \cot^{-1}:\ (0,\pi),\qquad \sin^{-1}:\ \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right].$ …

Common Mistakes

Mistake 1: Giving $\cot^{-1}\!\big(\tfrac{1}{\sqrt3}\big)$ a negative or reflex value.

Why it's wrong: $\cot^{-1}$ of a positive number is a first-quadrant angle in $(0,\pi)$; here it is $\tfrac{\pi}{3}$, always positive. Correct approach: keep $\cot^{-1}$ outputs in $(0,\pi)$ — do not treat $\cot^{-1}$ as odd like $\tan^{-1}$.

Mistake 2: Forgetting to simplify $\sin\!\big(-\tfrac{\pi}{2}\big)$ before applying $\tan^{-1}$.

Why it's wrong: leaving it symbolic invites errors; the inner value is simply $-1$. Correct approach: reduce first, then $\tan^{-1}(-1)=-\tfrac{\pi}{4}$. …