Example 5: Simplest form of $\cot^{-1}\frac{1}{\sqrt{x^2-1}}$

Basics you need

Solution

1. Check the expression is defined

   For $x>1$ we have $x^2>1$⁠, so $x^2 - 1 > 0$⁠. Its square root is a positive real number, so $\frac{1}{\sqrt{x^2-1}}$ is defined and positive. $\cot^{-1}$ accepts every real number, so $f(x)$ is defined for every $x>1$⁠.

   The condition $x>1$ does two jobs: it keeps the number under the root positive (so the root exists) and not zero (so we are not dividing by 0). At $x = 1$ the fraction would be $\frac{1}{0}$⁠.

2. Substitute $x = \sec\theta$ and locate $\theta$

   Let $\theta = \sec^{-1}x$⁠. This exists, because $x>1$ is in the domain $\R-(-1,1)$⁠. Then $\sec\theta = x$⁠, so $\cos\theta = \frac{1}{x}$⁠, and since $x>1$⁠,

   $$0 < \cos\theta < 1.$$

   In the range $[0,\pi]-\left\{\frac{\pi}{2}\right\}$⁠, cosine is positive only on $\left[0, \frac{\pi}{2}\right)$⁠, and $\cos\theta < 1$ rules out $\theta = 0$⁠. Therefore

   $$0 < \theta < \frac{\pi}{2}.$$
   The textbook says "Let $x = \sec\theta$" and at the end writes $\theta = \sec^{-1}x$⁠. That last equality is only true if $\theta$ is in the principal range of $\sec^{-1}$⁠, so we define $\theta$ that way from the start. Knowing that $\theta$ is acute is the fact that Steps 3 and 4 rely on.
   On the unit circle, angles in $[0,\pi]$ with positive cosine are those to the right of the $y$-axis, $0\le\theta<\frac{\pi}{2}$⁠. Lesson: principal values

3. Simplify the square root, with the correct sign

   $$\begin{aligned} \sqrt{x^2-1} &= \sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} \\ &= |\tan\theta| = \tan\theta. \end{aligned}$$

   The last equality holds because $\tan\theta>0$ for $0<\theta<\frac{\pi}{2}$ (both $\sin\theta$ and $\cos\theta$ are positive in the first quadrant). Hence

   $$\frac{1}{\sqrt{x^2-1}} = \frac{1}{\tan\theta} = \cot\theta.$$
   The textbook writes $\sqrt{\sec^2\theta-1} = \tan\theta$ directly. That is only true when $\tan\theta\ge0$⁠, which is why Step 2 matters. (If $\tan\theta$ were negative, the square root would be $-\tan\theta$⁠.) Since $\tan\theta>0$⁠, the reciprocal $\cot\theta$ exists and is positive too.
   $\sec^2\theta - 1 = \tan^2\theta$ (Basic 2) and $\sqrt{a^2}=|a|$ (Basic 3).

4. Cancel $\cot^{-1}(\cot\theta)$⁠, after checking $\theta$

   Now $f(x) = \cot^{-1}(\cot\theta)$⁠. Since $0<\theta<\frac{\pi}{2}$⁠, certainly $0<\theta<\pi$⁠, so $\theta$ is in the principal range of $\cot^{-1}$ and

   $$\begin{aligned} \cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right) &= \cot^{-1}(\cot\theta) \\ &= \theta = \sec^{-1}x. \end{aligned}$$
   This cancellation is the reason the problem reduces to a single inverse function. It is valid because $\theta$ is in $(0,\pi)$⁠; the final replacement $\theta = \sec^{-1}x$ is just the definition from Step 2.
   $\cot^{-1}(\cot t) = t$ for $t\in(0,\pi)$ (Basic 4). Lesson: substitutions and simplest form

5. See it in a right triangle

   Because $\theta$ is acute, we can draw it in a right triangle. Take the side adjacent to $\theta$ to be 1 and the hypotenuse to be $x$ (possible, since $x>1$⁠). By Pythagoras the opposite side is $\sqrt{x^2-1}$⁠. Reading off the ratios:

   $$\sec\theta = \frac{x}{1} = x, \qquad \cot\theta = \frac{1}{\sqrt{x^2-1}}.$$

   So the angle whose cotangent is $\frac{1}{\sqrt{x^2-1}}$ is the same acute angle whose secant is $x$⁠, which is the result of Step 4.

   The triangle is a quick way to discover or check such results, and it shows the answer could equally be written $\tan^{-1}\sqrt{x^2-1}$ or $\cos^{-1}\frac{1}{x}$⁠. The algebra of Steps 2 to 4 is still needed, because the triangle only makes sense for acute angles, and it is Step 2 that proves $\theta$ is acute.
   Pythagoras: $1^2 + \left(\sqrt{x^2-1}\right)^2 = x^2$⁠. Lesson: the triangle trick

Every fact used, and where it comes from