Q.If $\sin^{-1}\left(\frac{2a}{1+a^2}\right)+\cos^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, where $a, x\in\,]0,1[$, then the value of $x$ is
(A) $0$
(B) $\frac{a}{2}$
(C) $a$
(D) $\frac{2a}{1-a^2}$

Concept: Standard inverse trigonometric substitutions — for $a \in (0,1)$, let $a = \tan\theta$ so that $\theta \in (0,\pi/4)$. Then simplify each term using the double-angle identities.

Let $a = \tan\theta$, $\theta \in (0,\pi/4)$. Then:

• $\frac{2a}{1+a^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$, so $\sin^{-1}(\sin 2\theta) = 2\theta$ (since $2\theta \in (0,\pi/2)$).
• $\frac{1-a^2}{1+a^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$, so $\cos^{-1}(\cos 2\theta) = 2\theta$ (since $2\theta \in (0,\pi/2)$).

Thus the left side becomes $2\theta + 2\theta = 4\theta$. …

The core idea here is that when you see expressions like $\frac{2a}{1+a^2}$ or $\frac{1-a^2}{1+a^2}$, your mind should immediately jump to the tangent half-angle substitution. For $a \in (0,1)$, we can set $a = \tan\theta$ where $\theta \in (0, \pi/4)$. This turns those messy rational forms into clean trigonometric functions.

Why does this work? Because:

• $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2a}{1+a^2}$
• $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \frac{1-a^2}{1+a^2}$

Similarly, for $x \in (0,1)$, set $x = \tan\phi$ with $\phi \in (0, \pi/4)$, giving $\tan(2\phi) = \frac{2x}{1-x^2}$.

Now the inverse trig functions become straightforward: $\sin^{-1}(\sin 2\theta) = 2\theta$ and $\cos^{-1}(\cos 2\theta) = 2\theta$, because $2\theta \in (0, \pi/2)$ — well within the principal ranges of both functions. And $\tan^{-1}(\tan 2\phi) = 2\phi$ since $2\phi \in (0, \pi/2)$.

Let's work through it step by step.

1. Substitute $a = \tan\theta$. Since $a \in (0,1)$, we have $\theta \in (0, \pi/4)$. Then:

Both $2\theta$ lies in $(0, \pi/2)$, so the principal values of $\sin^{-1}$ and $\cos^{-1}$ give:

2. Substitute $x = \tan\phi$. With $x \in (0,1)$, we get $\phi \in (0, \pi/4)$. Then:

Since $2\phi \in (0, \pi/2)$, the principal value is:

3. Rewrite the given equation. The original equation: …

Method: The $t=\tan\theta$ substitution for standard rational arguments

Expressions $\dfrac{2t}{1+t^2}$, $\dfrac{1-t^2}{1+t^2}$, and $\dfrac{2t}{1-t^2}$ are the sine, cosine, and tangent of $2\theta$ when $t=\tan\theta$. Substituting collapses inverse-trig equations into linear ones in $\theta$.

Steps

Step 1: Substitute a tangent for each free variable.

Set $a=\tan\theta$ (and $x=\tan\phi$). Fix the range: since $a\in(0,1)$, $\theta\in\left(0,\tfrac{\pi}{4}\right)$; likewise for $\phi$. Tracking these ranges is essential for Step 3.

Step 2: Rewrite the rational arguments as double-angle values.

Apply

Step 3: Peel off the inverse functions — but only if the angle is in range. …

Common Mistakes

Mistake 1: Peeling off the inverse functions without checking the angle's range.

Why it's wrong: $\sin^{-1}(\sin 2\theta)=2\theta$ (and similarly for $\cos^{-1}$) only holds while $2\theta$ stays inside the principal range; otherwise a $\pi-2\theta$ correction is needed. Correct approach: the domain $a,x\in(0,1)$ keeps $2\theta,2\phi\in\left(0,\tfrac{\pi}{2}\right)$, so peeling off is valid.

Mistake 2: Miscounting the left side as $2\theta$ instead of $4\theta$.

Why it's wrong: both $\sin^{-1}$ and $\cos^{-1}$ terms equal $2\theta$, so they add to $4\theta$. Correct approach: $4\theta=2\phi\Rightarrow\phi=2\theta\Rightarrow x=\tan 2\theta=\tfrac{2a}{1-a^2}$. …