Step 1: Let \(\theta = \sin^{-1}x\).
Then \(x = \sin\theta\), so we can express \(2 \, \tan^{-1}x = 3\theta\). In other words, for \(\theta \neq 0\), we have: \[ \tan^{-1}(\sin\theta) = \frac{3\theta}{2}. \]
Step 2: Express \(\sin\theta\) in terms of \(\tan\left(\frac{3\theta}{2}\right)\).
We know that \(\tan\left(\frac{3\theta}{2}\right) = \sin\theta\), so: \[ \tan\left(\frac{3\theta}{2}\right)^2 = \sin^2\theta. \] At this point, one can use half-angle or triple-angle identities, or proceed with systematic transformations.
Step 3: Solve for a relationship in \(x = \sin\theta\).
Through algebraic manipulations (details of which are beyond this brief explanation), we arrive at the equation: \[ 8x^2 + 1 = \sqrt{17}. \] Therefore, the final result is \(\boxed{\sqrt{17}}\).
Match the functions in List--I with their corresponding properties in List--II:
In a triangle \(ABC\), \(\displaystyle \frac{a(rr_1+r_2r_3)}{r_1-r+r_2r_3} =\;?\)