Q.Find the principal value of the following: $\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}$, $|x| < a$

Concept: Inverse Trigonometric Graphs — Principal value by substitution

We simplify the expression inside the inverse tangent using a trigonometric substitution.

Let $x = a \sin \theta$, where $|x| < a$ implies $|\theta| < \frac{\pi}{2}$ (principal range of $\sin^{-1}$). Then:

Since $\cos \theta > 0$ for $|\theta| < \frac{\pi}{2}$, the square root is positive. …

We are asked to find the principal value of $\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}$, with the condition $|x| < a$. This is not a numeric evaluation — it's a simplification into a standard inverse trigonometric form. The key is to recognise the algebraic structure inside the inverse tangent.

The expression $\frac{x}{\sqrt{a^2-x^2}}$ looks like a ratio of two sides of a right triangle. If we set up a triangle where the opposite side is $x$ and the adjacent side is $\sqrt{a^2-x^2}$, then the hypotenuse becomes $a$ (by Pythagoras). That means the angle whose tangent is that ratio is also the angle whose sine is $x/a$.

Let's walk through this carefully.

1. Set up a right triangle. Consider a right triangle with angle $\theta$. Let the side opposite $\theta$ be $x$, and the side adjacent to $\theta$ be $\sqrt{a^2-x^2}$. Then:

So $\theta = \tan^{-1} \frac{x}{\sqrt{a^2-x^2}}$.

2. Find the hypotenuse. By the Pythagorean theorem:

Since $|x| < a$, $a$ could be positive or negative. But the expression $\sqrt{a^2-x^2}$ uses the principal square root, which is non-negative. For the triangle to make sense geometrically, we take $a > 0$ (or treat $a$ as a positive constant). In most exam contexts, $a$ is assumed positive unless stated otherwise. We'll proceed with $a > 0$.

3. Express $\sin \theta$. From the same triangle:

Hence $\theta = \sin^{-1} \frac{x}{a}$.

4. Check the principal value range. …

Method: Converting an inverse tangent to an inverse sine by substitution

Use this when a $\tan^{-1}$ argument contains $\sqrt{a^2 - x^2}$.

Steps

Step 1: Substitute $x = a\sin\theta$

With $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\sqrt{a^2 - x^2} = a\cos\theta$ (non-negative on this branch).

Step 2: Simplify the ratio …

Common Mistakes

Mistake 1: Simplifying $\sqrt{a^2 - x^2}$ to $a\cos\theta$ without noting $\cos\theta \ge 0$

Why it's wrong: the root is non-negative, valid only because $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ keeps $\cos\theta \ge 0$. Correct approach: state the branch of the substitution. …