Exercise 2.2 · Q15 (MCQ): $\tan^{-1}\sqrt3-\cot^{-1}\left(-\sqrt3\right)$

Basics you need

Solution

1. Find $\tan^{-1}\sqrt3$

   We need $y \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ with $\tan y = \sqrt3$⁠. Since $\tan\frac{\pi}{3} = \frac{\sin(\pi/3)}{\cos(\pi/3)}$⁠, which is $\frac{\sqrt3/2}{1/2} = \sqrt3$⁠, and $\frac{\pi}{3}$ is in the range,

   $$\tan^{-1}\sqrt3 = \frac{\pi}{3}.$$
   The input is positive, so the angle is in the first quadrant, and $\frac{\pi}{3}$ (60°) is the special angle with tangent $\sqrt3$⁠. By Basic 4 there is no other answer in the range.
   Principal value: right ratio and inside the principal range. Lesson: principal values

2. Find $\cot^{-1}\left(-\sqrt3\right)$⁠: mind the range

   We need $y \in (0, \pi)$ with $\cot y = -\sqrt3$⁠. A negative cotangent in $(0,\pi)$ means the second quadrant (there $\sin > 0$ and $\cos < 0$⁠). The reference angle is $\frac{\pi}{6}$⁠, because $\cot\frac{\pi}{6} = \sqrt3$⁠. So take $y = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$⁠:

   $$\cot\frac{5\pi}{6} = \frac{\cos(5\pi/6)}{\sin(5\pi/6)} = \frac{-\sqrt3/2}{1/2} = -\sqrt3.$$

   Since $0 < \frac{5\pi}{6} < \pi$⁠, and $\cot$ takes each value once on $(0,\pi)$⁠,

   $$\cot^{-1}\left(-\sqrt3\right) = \frac{5\pi}{6}.$$
   This is the step the question is built to test. Because $\cot\frac{\pi}{6} = \sqrt3$ and $\cot$ is odd, $\cot\left(-\frac{\pi}{6}\right) = -\sqrt3$ too, so $-\frac{\pi}{6}$ looks like an answer. But $-\frac{\pi}{6}$ is not in $(0,\pi)$⁠, so it is not a value of $\cot^{-1}$⁠. Unlike $\sin^{-1}$ and $\tan^{-1}$⁠, the function $\cot^{-1}$ has a range that is not symmetric about 0, so $\cot^{-1}(-x) \ne -\cot^{-1}x$⁠. The correct rule is $\cot^{-1}(-x) = \pi - \cot^{-1}x$⁠, which here gives $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$⁠.
   Why $\cot^{-1}(-x) = \pi - \cot^{-1}x$⁠: if $\theta = \cot^{-1}x \in (0,\pi)$⁠, then $\pi - \theta$ is also in $(0,\pi)$ and $\cot(\pi - \theta) = -\cot\theta = -x$ (Basic 3). Basics: tan, cot, sec and cosec

3. Subtract

   $$\begin{aligned} \tan^{-1}\sqrt3 - \cot^{-1}\left(-\sqrt3\right) &= \frac{\pi}{3} - \frac{5\pi}{6} \\ &= \frac{2\pi - 5\pi}{6} = -\frac{3\pi}{6} \\ &= -\frac{\pi}{2}. \end{aligned}$$
   Use the common denominator 6. A negative result is perfectly possible: the expression is a difference of two angles, not itself a principal value, so it need not lie in any principal range. (With $\tan^{-1}$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $\cot^{-1}$ in $(0,\pi)$⁠, such a difference can be anywhere in $\left(-\frac{3\pi}{2}, \frac{\pi}{2}\right)$⁠.)
   $\frac{\pi}{3} = \frac{2\pi}{6}$⁠. Basics: angles and radians

4. Examine every option

   Option	Verdict
   (A) $\pi$	Wrong. The value is $-\frac{\pi}{2}$⁠. $\pi$ would need the two angles to differ by $\pi$⁠, which would mean $\cot^{-1}\left(-\sqrt3\right) = -\frac{2\pi}{3}$⁠, outside $(0,\pi)$ and with $\cot\left(-\frac{2\pi}{3}\right) = \frac{1}{\sqrt3}$⁠, not $-\sqrt3$⁠.
   (B) $-\frac{\pi}{2}$	Correct, by Steps 1 to 3.
   (C) $0$	Wrong. That would need $\cot^{-1}\left(-\sqrt3\right) = \frac{\pi}{3}$⁠, but $\cot\frac{\pi}{3} = \frac{1}{\sqrt3}$⁠, which is positive.
   (D) $2\sqrt3$	Wrong. It treats $\tan^{-1}$ and $\cot^{-1}$ as if they did nothing, computing $\sqrt3 - (-\sqrt3)$⁠. The values of inverse functions are angles, and the answer is an angle.

   So the answer is (B).

   Note that the most tempting wrong value, from $\cot^{-1}\left(-\sqrt3\right) = -\frac{\pi}{6}$⁠, is $\frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$⁠. It is not among the options, but $-\frac{\pi}{2}$ is, and a student who drops a sign at the end could land on (B) for the wrong reason. Only the full argument of Step 2 makes (B) safe.

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