Q.Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1} \left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$

Step 1: Let $x = \tan\theta$. Since $0 < x < 1$, $\theta \in \left(0, \tfrac{\pi}{4}\right)$. Then

$\frac{1-x^2}{1+x^2} = \cos 2\theta, \qquad y = \sin^{-1}(\cos 2\theta) = \sin^{-1}\!\left[\sin\!\left(\frac{\pi}{2}-2\theta\right)\right].$ …

Recognising the Identity

As in the neighbouring $\cos^{-1}$ version of this expression, $\dfrac{1-x^2}{1+x^2} = \cos 2\theta$ under $x = \tan\theta$. Here the outer function is $\sin^{-1}$, so $\cos 2\theta$ must first be rewritten as a sine using the complementary-angle identity before the inverse sine can cancel it.

Step-by-Step Solution

1. Substitute $x = \tan\theta$.

Since $0 < x < 1$, let $\theta = \tan^{-1}x \in \left(0, \tfrac{\pi}{4}\right)$. Then:

2. Rewrite the cosine as a sine.

Using $\cos\phi = \sin\!\left(\tfrac{\pi}{2}-\phi\right)$ with $\phi = 2\theta$:

3. Check the range.

Since $\theta \in \left(0, \tfrac{\pi}{4}\right)$, $2\theta \in \left(0, \tfrac{\pi}{2}\right)$, so:

$\frac{\pi}{2}-2\theta \in \left(0, \frac{\pi}{2}\right) \subset \left[-\frac{\pi}{2}, \frac{\pi}{2}\right],$ …

Method: Converting Between Inverse-Sine and Inverse-Cosine Forms via the Complementary Identity

This method is needed when the double-angle substitution naturally produces a cosine (like $\cos 2\theta$), but the outer inverse function is $\sin^{-1}$, not $\cos^{-1}$ — you must convert cosine to sine using the complementary-angle identity before you can cancel.

Steps

Step 1: Substitute and identify which trig function you land on

With $x=\tan\theta$, $\frac{1-x^2}{1+x^2}=\cos 2\theta$, so the expression is $y=\sin^{-1}(\cos 2\theta)$ — a cosine trapped inside an inverse sine.

Step 2: Rewrite the cosine as a sine using the complementary identity

Applying this with $\phi=2\theta$ gives $\cos 2\theta = \sin\!\left(\frac{\pi}{2}-2\theta\right)$, so $y=\sin^{-1}\!\left[\sin\!\left(\frac{\pi}{2}-2\theta\right)\right]$ — now the inner and outer functions match.

Step 3: Check the new angle against $\sin^{-1}$'s range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ …

Common Mistakes

Mistake 1: Applying $\sin^{-1}(\cos\alpha)=\alpha$ directly, without converting

Because the inner expression naturally simplifies to $\cos 2\theta$, it's tempting to plug it straight into $\sin^{-1}$ and cancel, as if $\sin^{-1}(\cos\alpha)=\alpha$ were a valid identity. Why it's wrong: $\sin^{-1}$ only undoes $\sin$, not $\cos$ — you must first rewrite $\cos 2\theta$ as $\sin\left(\frac{\pi}{2}-2\theta\right)$ using the complementary-angle identity before the inverse sine can cancel it. Correct approach: always match the trig function inside the expression to the SAME trig function as the outer inverse before cancelling; convert first if they don't match.

Mistake 2: Mishandling the sign when converting via the complementary identity …