Q.Find the principal value of the following: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$, $x \neq 0$

Concept: Inverse Trigonometric Graphs — the principal value branch of $\tan^{-1}$ is $(-\pi/2, \pi/2)$, so any expression must be reduced to an angle in that interval.

Step 1: Let $x = \tan \theta$, where $\theta \in (-\pi/2, \pi/2)$. Then $\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = |\sec\theta|$. Since $\theta$ is in $(-\pi/2, \pi/2)$, $\sec\theta > 0$, so $|\sec\theta| = \sec\theta$.

Step 2: The expression becomes

Step 3: Using the identity $\frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2}$, we get

Since $\theta \in (-\pi/2, \pi/2)$, we have $\theta/2 \in (-\pi/4, \pi/4)$, which lies inside the principal branch of $\tan^{-1}$. Hence the value is $\theta/2 = \frac{1}{2}\tan^{-1}x$.

Why Inverse Trigonometric Graphs Matter Here

When you see an expression like $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$, your first instinct might be to try algebraic simplification directly. That works, but it’s messy. The cleaner path is to recognise that $\sqrt{1+x^2}$ screams for a trigonometric substitution — specifically, $x = \tan \theta$. Why? Because $1 + \tan^2 \theta = \sec^2 \theta$, and the square root becomes $|\sec \theta|$, which is much friendlier.

The key insight: inverse trigonometric functions are angles. So $\tan^{-1}(\text{something})$ is asking: what angle has this tangent? If we can rewrite the “something” as the tangent of a simpler angle, we’re done.

Let’s walk through it.

1. Set up the substitution

   Let $x = \tan \theta$, where $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ — the principal branch of $\tan^{-1}$. Then $\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = |\sec \theta|$.

   Since $\theta$ is in $(-\pi/2, \pi/2)$, $\sec \theta > 0$, so $|\sec \theta| = \sec \theta$.

   Thus the expression becomes:

2. Rewrite in terms of sine and cosine $\sec \theta = \frac{1}{\cos \theta}$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$. So:

3. Use the half-angle identity Recall: $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$. So:

4. Back-substitute We now have:

But $\theta = \tan^{-1} x$, so $\frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$.

Now, is $\frac{1}{2} \tan^{-1} x$ always in the principal range of $\tan^{-1}$, i.e., $(-\pi/2, \pi/2)$?

Since $\tan^{-1} x \in (-\pi/2, \pi/2)$, half of it lies in $(-\pi/4, \pi/4)$, which is safely inside $(-\pi/2, \pi/2)$. So the identity $\tan^{-1}(\tan \alpha) = \alpha$ holds for $\alpha = \frac{1}{2} \tan^{-1} x$.

Therefore:

Method: Simplifying an inverse tangent by trig substitution

Use this when a $\tan^{-1}$ argument contains $\sqrt{1+x^2}$ (or $\sqrt{a^2 - x^2}$, $\sqrt{x^2 - a^2}$).

Steps

Step 1: Choose the substitution that removes the radical

For $\sqrt{1+x^2}$, set $x = \tan\theta$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $\sqrt{1+x^2} = \sec\theta$ (positive on this branch).

Step 2: Rewrite everything in $\sin\theta$ and $\cos\theta$

Replace $\sec\theta$ and $\tan\theta$, then simplify the fraction to a recognisable half-angle form such as $\frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}$.

Step 3: Apply the inverse and back-substitute

$\tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2}$ provided $\frac{\theta}{2}$ is in the branch (it is, since $\frac{\theta}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$). Replace $\theta = \tan^{-1}x$ to give the answer in $x$.

Common Mistakes

Mistake 1: Taking $\sqrt{1+x^2} = \sqrt{\sec^2\theta} = \sec\theta$ without justifying the sign

Why it's wrong: in general $\sqrt{\sec^2\theta} = |\sec\theta|$; dropping to $\sec\theta$ is valid only because $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ makes $\sec\theta > 0$. Correct approach: state the branch to remove the absolute value.

Mistake 2: Forgetting the half-angle and answering $\tan^{-1}x$

Why it's wrong: the simplification produces $\frac{\theta}{2}$, so the result is $\frac{1}{2}\tan^{-1}x$, not $\tan^{-1}x$. Correct approach: track the factor of $\frac{1}{2}$ from $\tan\frac{\theta}{2}$.