Exercise 2.2 · Q3: Simplest form of $\tan^{-1}\frac{\sqrt{1+x^2}-1}{x}$

Basics you need

Solution

1. Substitute $x = \tan\theta$

   Let $\theta = \tan^{-1}x$⁠. This is defined for every real $x$⁠, and

   $$\tan\theta = x, \qquad \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$$

   Since $x \ne 0$⁠, also $\theta \ne 0$⁠.

   $1 + x^2$ under a root is the pattern $1 + \tan^2\theta = \sec^2\theta$⁠. Using $\theta = \tan^{-1}x$ (not any other angle with tangent $x$⁠) gives us its interval, which we need in the next step and in Step 5.
   $\tan(\tan^{-1}x) = x$ for all real $x$⁠. Lesson: substitutions and simplest form

2. Remove the square root, with the right sign

   $$\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = \sqrt{\sec^2\theta} = |\sec\theta|.$$

   For $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ we have $\cos\theta \gt 0$⁠, so $\sec\theta \gt 0$ and $|\sec\theta| = \sec\theta$⁠. Therefore

   $$\frac{\sqrt{1+x^2}-1}{x} = \frac{\sec\theta - 1}{\tan\theta}.$$
   This is where the principal range first matters. If $\theta$ were, say, between $\frac{\pi}{2}$ and $\pi$⁠, then $\sec\theta$ would be negative and $\sqrt{\sec^2\theta}$ would be $-\sec\theta$⁠. Writing $\sec\theta$ there would be wrong.
   $\sqrt{a^2} = |a|$ (Basic 2); cosine is positive in the first and fourth quadrants. Basics: sin and cos on the unit circle

3. Write everything with sin and cos

   Multiply top and bottom by $\cos\theta$ (which is not 0):

   $$\frac{\sec\theta - 1}{\tan\theta} = \frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}} = \frac{1 - \cos\theta}{\sin\theta}.$$

   The denominator $\sin\theta$ is not 0, because $\theta \ne 0$ and $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠.

   Clearing the small fractions leaves a form in which a standard identity is recognisable. Checking that we never divide by zero is part of the job: $\cos\theta \ne 0$ by the interval, $\sin\theta \ne 0$ because $x \ne 0$⁠. This is exactly why the question says $x \ne 0$⁠.

4. Derive the half-angle forms and simplify

   Put $u = \frac{\theta}{2}$⁠, so $\theta = 2u$⁠. The double-angle formulas give

   $$1 - \cos\theta = 1 - (1 - 2\sin^2 u) = 2\sin^2\frac{\theta}{2},$$ $$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.$$

   Hence

   $$\begin{aligned} \frac{1-\cos\theta}{\sin\theta} &= \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \\ &= \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}. \end{aligned}$$

   The cancellation is allowed: $\frac{\theta}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ and $\frac{\theta}{2} \ne 0$⁠, so neither $\sin\frac{\theta}{2}$ nor $\cos\frac{\theta}{2}$ is 0.

   The "1 minus cos" in the numerator is the clue: $\cos\theta = 1 - 2\sin^2\frac{\theta}{2}$ is the version of the double-angle formula that makes the 1 cancel. Writing $\sin\theta$ also in terms of $\frac{\theta}{2}$ lets a common factor $2\sin\frac{\theta}{2}$ cancel.
   $\cos 2u = 1 - 2\sin^2 u$⁠, $\sin 2u = 2\sin u\cos u$ (Basic 4), Class XI trigonometry.

5. Check the principal range and conclude

   The function is now $\tan^{-1}\left(\tan\frac{\theta}{2}\right)$⁠. Since $-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$⁠,

   $$-\frac{\pi}{4} \lt \frac{\theta}{2} \lt \frac{\pi}{4},$$

   which lies inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. So $\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2}$⁠, and

   $$\tan^{-1}\frac{\sqrt{1+x^2}-1}{x} = \frac{\theta}{2} = \frac12\tan^{-1}x.$$

   This holds for every real $x \ne 0$⁠. There is no further restriction, because halving an angle of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ can never leave that interval.

   Unlike Q1 and Q2, nothing goes wrong anywhere in the stated domain. The only excluded point is $x = 0$⁠, where the expression is $\frac{0}{0}$ and is simply undefined (the answer $\frac12\tan^{-1}0 = 0$ is its limiting value there).
   $\tan^{-1}(\tan y) = y$ for $y\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Lesson: sin⁻¹(sin x) and the folding rules

6. An algebra cross-check

   Multiply top and bottom by $\sqrt{1+x^2} + 1$⁠:

   $$\begin{aligned} \frac{\sqrt{1+x^2}-1}{x} &= \frac{(1+x^2) - 1}{x\left(\sqrt{1+x^2}+1\right)} \\ &= \frac{x}{\sqrt{1+x^2}+1}. \end{aligned}$$

   The denominator is greater than $|x|$⁠, so this number lies strictly between $-1$ and $1$ and has the same sign as $x$⁠. So $\tan^{-1}$ of it lies in $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$⁠, matching the range of $\frac12\tan^{-1}x$⁠.

   A second, independent route catches mistakes. Here it confirms the sign and the size of the answer without any trigonometry.
   Rationalising: $(a-b)(a+b) = a^2 - b^2$ removes a square root from a numerator.

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