Solving the Given Trigonometric Equation

We are asked to find the value of \(x\) in the given trigonometric equation:

\(\cos^{-1} \dfrac{1-a^2}{1 + a^2} - \cos^{-1} \dfrac{1-b^2}{1 + b^2} = 2 \tan^{-1} x\)

This problem involves inverse trigonometric functions and can be simplified using standard trigonometric identities. The forms \(\dfrac{1-t^2}{1+t^2}\) are strong indicators that we should use substitutions related to the tangent function.

Applying Trigonometric Identities

Let's use the substitution method to simplify the terms involving \(a\) and \(b\). We know the identity \(\cos(2\theta) = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}\). This identity relates the cosine of a double angle to the tangent of the angle.

Let \(a = \tan \alpha\) and \(b = \tan \beta\). Assuming that \(a\) and \(b\) are within the domains where \(\alpha = \tan^{-1} a\) and \(\beta = \tan^{-1} b\) are well-defined and the relevant principal values of inverse cosine are used, the equation becomes:

\(\cos^{-1} \left(\dfrac{1-\tan^2\alpha}{1 + \tan^2\alpha}\right) - \cos^{-1} \left(\dfrac{1-\tan^2\beta}{1 + \tan^2\beta}\right) = 2 \tan^{-1} x\)

Using the identity \(\cos(2\theta) = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}\), we can rewrite the equation as:

\(\cos^{-1}(\cos(2\alpha)) - \cos^{-1}(\cos(2\beta)) = 2 \tan^{-1} x\)

Simplifying with Inverse Trigonometric Functions

For appropriate ranges of \(\alpha\) and \(\beta\), we have \(\cos^{-1}(\cos(2\alpha)) = 2\alpha\) and \(\cos^{-1}(\cos(2\beta)) = 2\beta\). Assuming these principal values, the equation simplifies to:

\(2\alpha - 2\beta = 2 \tan^{-1} x\)

Dividing both sides by 2, we get:

\(\alpha - \beta = \tan^{-1} x\)

Substituting Back and Solving for x

Now, substitute back the original expressions for \(\alpha\) and \(\beta\). Since we set \(a = \tan \alpha\) and \(b = \tan \beta\), it follows that \(\alpha = \tan^{-1} a\) and \(\beta = \tan^{-1} b\).

Substituting these back into the simplified equation:

\(\tan^{-1} a - \tan^{-1} b = \tan^{-1} x\)

Next, we use another important identity involving inverse trigonometric functions: \(\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left(\dfrac{A - B}{1 + AB}\right)\), provided \(AB > -1\).

Applying this identity to the left side of our equation:

\(\tan^{-1} \left(\dfrac{a - b}{1 + ab}\right) = \tan^{-1} x\)

Since the \(\tan^{-1}\) function is a one-to-one function, if \(\tan^{-1} P = \tan^{-1} Q\), then \(P = Q\). Therefore, comparing the arguments of the \(\tan^{-1}\) functions on both sides, we find the value of \(x\):

\(x = \dfrac{a - b}{1 + ab}\)

This is the value of \(x\) that satisfies the given trigonometric equation under the assumed conditions for the inverse trigonometric functions and identities. Solving equations like this demonstrates the power of using trigonometric identities to simplify complex expressions.

The value of \(x\) obtained by solving this trigonometric equation is \(\dfrac{a - b}{1 + ab}\).