Q.$\sin^{-1} (1-x) - 2 \sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to (A) $0, \frac{1}{2}$ (B) $1, \frac{1}{2}$ (C) $0$ (D) $\frac{1}{2}$

Concept: Inverse Tangent Identity — When an equation mixes $\sin^{-1}$ terms and yields a constant like $\frac{\pi}{2}$, rewriting one side using $\sin^{-1} a = \frac{\pi}{2} - \sin^{-1} \sqrt{1-a^2}$ (for $a \ge 0$) often simplifies the work.

Step 1:

Given $\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}$.

Bring $2\sin^{-1} x$ to the right:

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

Step 2:

Take sine of both sides. Since $\sin(\frac{\pi}{2} + \theta) = \cos\theta$, we get:

$1-x = \cos(2\sin^{-1} x)$.

Step 3:

Let $\theta = \sin^{-1} x$, so $\sin\theta = x$ and $\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2x^2$.

Thus: $1-x = 1 - 2x^2 \implies -x = -2x^2 \implies 2x^2 - x = 0$.

Step 4: …

Concept and Intuition

When an inverse trigonometric equation involves a mix of $\sin^{-1}$ and a constant like $\frac{\pi}{2}$, a powerful trick is to convert one of the inverse sines into an inverse cosine using the complementary identity:

This lets us replace $\sin^{-1}(1-x)$ with $\frac{\pi}{2} - \cos^{-1}(1-x)$, which often simplifies the equation into a form where we can take sine of both sides cleanly. The alternative — taking sine directly — leads to messy algebra and risks missing domain restrictions. The complementary identity keeps the logic crisp and the domain checks natural.

Step-by-step solution

1. Apply the complementary identity

We know:

Substitute into the given equation:

Cancel $\frac{\pi}{2}$ from both sides:

So:

But $\cos^{-1}$ of anything is always between $0$ and $\pi$, while the right side is negative unless $\sin^{-1}x = 0$. This tells us something important — but let’s proceed algebraically and check domains at the end.

2. Take cosine of both sides

Since $\cos(\cos^{-1} u) = u$ for $u \in [-1,1]$, we get:

Cosine is even: $\cos(-\theta) = \cos\theta$, so:

3. Use the double-angle identity

Let $\theta = \sin^{-1}x$. Then $\sin\theta = x$, and $\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2x^2$.

Thus:

4. Solve the quadratic

Cancel $1$ from both sides:

Factor:

So $x = 0$ or $x = \frac12$.

5. Check domain and original equation …

Method: Solving an inverse-sine equation and screening extraneous roots

Use this for equations that mix several $\sin^{-1}$ terms equalling a constant.

Steps

Step 1: Isolate one inverse term.

Move the other inverse terms across so a single $\sin^{-1}$ sits alone, e.g. $\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1}x$.

Step 2: Apply sine (or cosine) to both sides. …

Common Mistakes

Mistake 1: Accepting both roots $x = 0$ and $x = \frac{1}{2}$.

Why it's wrong: taking sine of both sides can create extraneous solutions; testing $x = \frac{1}{2}$ gives $\frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \ne \frac{\pi}{2}$, so it fails. Correct approach: substitute each candidate into the original equation; only $x = 0$ works (option C).

Mistake 2: Ignoring the domain of $\sin^{-1}(1-x)$. …