Q12:

Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x > 0)$

Step 1: Identifying the Algebraic Transformation

The given equation is $\tan^{-1} \left( \frac{1-x}{1+x} \right) = \frac{1}{2} \tan^{-1} x$. We recognize the left-hand side as a standard identity form. Recall the formula for the difference of tangents: $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$.

Step 2: Decomposition using Trigonometric Identities

Rewrite the expression $\frac{1-x}{1+x}$ as $\frac{1-x}{1+(1)(x)}$. Applying the identity [$\tan^{-1} 1 - \tan^{-1} x = \tan^{-1} \left( \frac{1-x}{1+1 \cdot x} \right)$]:

$\tan^{-1}(1) - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x)$

Step 3: Solving the Linear Equation in terms of $\tan^{-1} x$

Substitute the principal value of $\tan^{-1}(1)$, which is $\frac{\pi}{4}$:

$\frac{\pi}{4} - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x)$

Add $\tan^{-1}(x)$ to both sides of the equation [By the Additive Property of Equality]:

$\frac{\pi}{4} = \frac{1}{2} \tan^{-1}(x) + \tan^{-1}(x)$

$\frac{\pi}{4} = \frac{3}{2} \tan^{-1}(x)$

Step 4: Isolating $x$

Multiply by $\frac{2}{3}$ to isolate $\tan^{-1}(x)$:

$\tan^{-1}(x) = \frac{\pi}{4} \cdot \frac{2}{3} = \frac{\pi}{6}$

Take the tangent of both sides [Definition of Inverse Trigonometric Functions]:

$x = \tan\left(\frac{\pi}{6}\right)$

$x = \frac{1}{\sqrt{3}}$

Step 5: Verification of Domain Constraint

The problem provides the condition $x > 0$. Since $\frac{1}{\sqrt{3}} \approx 0.577$, the result satisfies $x > 0$ [Condition Check].

Final Solution: x = \frac{1}{\sqrt{3}}