Q.Find the value of the following: $\tan \frac{1}{2}\left[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}\right]$, where $|x| < 1$, $y > 0$ and $xy < 1$.

Concept: Inverse Tangent Identity — Use the standard substitutions $x = \tan \theta$ and $y = \tan \phi$ to simplify the inverse trigonometric expressions.

Let $x = \tan \theta$, where $\theta \in (-\pi/4, \pi/4)$ since $|x| < 1$. Then

$\sin^{-1} \frac{2x}{1+x^2} = \sin^{-1}(\sin 2\theta) = 2\theta$.

Let $y = \tan \phi$, with $\phi \in (0, \pi/4)$ because $y > 0$. Then

$\cos^{-1} \frac{1-y^2}{1+y^2} = \cos^{-1}(\cos 2\phi) = 2\phi$. …

We start with the expression inside the brackets:

The key insight is that both terms are standard forms of the inverse tangent function. For $|x| < 1$, we have the identity

Similarly, for $y > 0$,

So the sum inside the brackets becomes

Now the original expression is

We now use the tangent addition formula:

Here $A = \tan^{-1}x$ and $B = \tan^{-1}y$, so $\tan A = x$, $\tan B = y$. Therefore …

Method: Using the $2\tan^{-1}$ conversion identities

Use this when arguments look like $\frac{2x}{1+x^2}$, $\frac{1-x^2}{1+x^2}$, or $\frac{2x}{1-x^2}$.

Steps

Step 1: Replace each composite with a $2\tan^{-1}$ form

For the stated domain, $\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x$ and $\cos^{-1}\frac{1-y^2}{1+y^2} = 2\tan^{-1}y$.

Step 2: Simplify inside the bracket …

Common Mistakes

Mistake 1: Misremembering which composite converts to which $2\tan^{-1}$ form

Why it's wrong: $\frac{2x}{1+x^2}$ pairs with $\sin^{-1}$ and $\frac{1-y^2}{1+y^2}$ with $\cos^{-1}$, each equal to $2\tan^{-1}$ of the variable only on the stated domain. Correct approach: match the pattern carefully and note $|x|<1$, $y>0$ keep the conversions valid.

Mistake 2: Forgetting the outer $\frac{1}{2}$ before taking the tangent …