Q.Prove that: $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right)$, $x \in [0, 1]$.

Concept: Inverse Tangent Identity — converting $\tan^{-1}\sqrt{x}$ into an inverse cosine form using a right-triangle substitution.

Let $\theta = \tan^{-1}\sqrt{x}$. Then $\tan\theta = \sqrt{x}$, so we can draw a right triangle with opposite side $\sqrt{x}$ and adjacent side $1$. The hypotenuse is $\sqrt{1 + x}$.

From this triangle:

We need $\cos(2\theta)$. Using the double-angle formula:

$\cos(2\theta) = 2\cos^2\theta - 1 = 2\left(\frac{1}{1+x}\right) - 1 = \frac{2}{1+x} - 1 = \frac{1-x}{1+x}.$ …

Why This Approach Works

The core idea is to use the inverse tangent identity — a relationship between $\tan \theta$ and $\cos 2\theta$. When you have $\tan^{-1} \sqrt{x}$, you can set $\theta = \tan^{-1} \sqrt{x}$, so $\tan \theta = \sqrt{x}$. Then, using the double-angle formula for cosine in terms of tangent, you can express $\cos 2\theta$ purely in terms of $x$. This directly gives $2\theta = \cos^{-1}(\text{something})$, which is exactly the form we need.

The trick is that the domain $x \in [0,1]$ ensures $\theta \in [0, \pi/4]$, so $2\theta \in [0, \pi/2]$, where $\cos^{-1}$ gives a principal value that matches.

Step-by-Step Proof

1. Set up the substitution.

Let $\theta = \tan^{-1} \sqrt{x}$. Then by definition, $\tan \theta = \sqrt{x}$, and since $x \in [0,1]$, we have $\sqrt{x} \in [0,1]$, so $\theta \in [0, \pi/4]$.

2. Recall the double-angle formula for cosine in terms of tangent.

A standard identity is:

This comes from $\cos 2\theta = \cos^2\theta - \sin^2\theta$, dividing numerator and denominator by $\cos^2\theta$.

3. Substitute $\tan \theta = \sqrt{x}$.

4. Relate $2\theta$ to the inverse cosine.

Since $\theta \in [0, \pi/4]$, we have $2\theta \in [0, \pi/2]$. On this interval, the cosine function is one-to-one and its inverse gives the principal value. Therefore: …

Method: Proving a $\tan^{-1}$-to-$\cos^{-1}$ identity via substitution

Use this to prove relations like $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\!\left(\frac{1-x}{1+x}\right)$.

Steps

Step 1: Substitute the inverse term as an angle.

Let $\theta = \tan^{-1}\sqrt{x}$, so $\tan\theta = \sqrt{x}$. The goal becomes showing $2\theta = \cos^{-1}\!\left(\frac{1-x}{1+x}\right)$.

Step 2: Express $\cos 2\theta$ through the tangent.

$\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1-x}{1+x}.$ …

Common Mistakes

Mistake 1: Using the wrong double-angle form for cosine.

Why it's wrong: the identity needed here is $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$; using $\cos 2\theta = 2\cos^2\theta - 1$ without expressing $\cos\theta$ in terms of $x$ leaves the algebra stuck. Correct approach: with $\tan\theta = \sqrt{x}$, the tangent form gives $\frac{1-x}{1+x}$ immediately.

Mistake 2: Ignoring the domain when applying $\cos^{-1}$. …