Question

Evaluate : $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) + \tan^{-1}\left(\tan\left(\frac{2\pi}{3}\right)\right)$}

Hint:

Remember that $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ holds true for identical values of $x$. However, here the inputs are $-\frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}}$, so evaluate them directly rather than misapplying the identity.

Answer 1

Concept: Evaluating an inverse trigonometric expression requires systematically simplifying each term within its standard principal value branch.
• $\tan^{-1}(x)$ principal branch is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\tan^{-1}(-x) = -\tan^{-1}(x)$.
• $\cot^{-1}(x)$ principal branch is $(0, \pi)$.
• For the term $\tan^{-1}(\tan \phi)$, if $\phi$ falls outside the principal value branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we must map it back into the range using trigonometric periodicity identities.

Step 1: Evaluate the first three terms of the given expression independently.
Let the given expression be denoted by $E$: \[ E = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) + \tan^{-1}\left(\tan\left(\frac{2\pi}{3}\right)\right) \] Let us evaluate each term step by step: * Term 1: $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$ * Term 2: $\cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$ * Term 3: Since $\sin\left(-\frac{\pi}{2}\right) = -1$, this term becomes: \[ \tan^{-1}(-1) = -\frac{\pi}{4} \]

Step 2: Evaluate the fourth term using principal value rules and combine all elements.
* Term 4: Consider $\tan^{-1}\left(\tan\left(\frac{2\pi}{3}\right)\right)$. Notice that $\frac{2\pi}{3}$ is not in the principal domain $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We rewrite it using the identity $\tan(\pi - \theta) = -\tan\theta$: \[ \tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = \tan\left(-\frac{\pi}{3}\right) \] Since $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we can simplify: \[ \tan^{-1}\left(\tan\left(-\frac{\pi}{3}\right)\right) = -\frac{\pi}{3} \] Now substitute all four evaluated values back into the expression $E$: \[ E = \left(-\frac{\pi}{6}\right) + \left(\frac{\pi}{3}\right) + \left(-\frac{\pi}{4}\right) + \left(-\frac{\pi}{3}\right) \] Notice that $+\frac{\pi}{3}$ and $-\frac{\pi}{3}$ cancel each other out cleanly: \[ E = -\frac{\pi}{6} - \frac{\pi}{4} \] Find a common denominator, which is 12: \[ E = \frac{-2\pi - 3\pi}{12} = -\frac{5\pi}{12} \] *Correction review check:* Let's double check the initial expression values: Term 1: $-\pi/6$. Term 2: $\pi/3$. Term 3: $-\pi/4$. Term 4: $-\pi/3$. Sum: $-\pi/6 + \pi/3 - \pi/4 - \pi/3 = -\pi/6 - \pi/4 = -5\pi/12$.