Q.Solve the following equation: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1} x$, $(x > 0)$.

Concept: Inverse Tangent Identity

We use the identity $\tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ and the fact that $\tan^{-1}(1) = \frac{\pi}{4}$.

Step 1: Rewrite the left side.

Notice $\frac{1-x}{1+x} = \frac{1 - x}{1 + x} = \frac{1 - x}{1 + 1\cdot x}$. This matches $\tan^{-1}(1) - \tan^{-1}(x)$ because

Step 2: So the equation becomes

Since $\tan^{-1}1 = \frac{\pi}{4}$, we have …

We start with the equation

The left side looks like the formula for $\tan^{-1}1 - \tan^{-1}x$. Recall the identity:

Here, take $a = 1$ and $b = x$. Then

Since $\tan^{-1}1 = \frac{\pi}{4}$, the equation becomes

1. Combine the inverse tangent terms. Bring $\tan^{-1}x$ terms together:

So

2. Take the tangent of both sides. Since $x > 0$, the principal value of $\tan^{-1}x$ lies in $(0, \frac{\pi}{2})$, so we can safely apply $\tan$:

3. Check the domain and validity. …

Method: Solving an equation by recognising a subtraction identity

Use this when an inverse tangent contains $\frac{1-x}{1+x}$ (or a similar $\frac{a-b}{1+ab}$ shape).

Steps

Step 1: Rewrite the fraction as a difference of inverse tangents.

Because $\tan^{-1}1 - \tan^{-1}x = \tan^{-1}\!\left(\frac{1-x}{1+x}\right)$ for $x > -1$, and $\tan^{-1}1 = \frac{\pi}{4}$:

Step 2: Reduce to a linear equation in $\tan^{-1}x$. …

Common Mistakes

Mistake 1: Using the difference identity without checking its condition.

Why it's wrong: $\tan^{-1}1 - \tan^{-1}x = \tan^{-1}\frac{1-x}{1+x}$ requires $1\cdot x > -1$; for $x > 0$ this holds, but blindly applying it for negative $x$ can be wrong. Correct approach: confirm $x > 0$, then rewrite the left side as $\frac{\pi}{4} - \tan^{-1}x$.

Mistake 2: Mishandling the linear step, giving $\tan^{-1}x = \frac{\pi}{4}$. …