Question

If the value of x satisfying the equation $\sin^{-1}\sqrt{1-x^2}=\tan^{-1}\frac{\sqrt{2-1}}{x}$ is $\frac{a}{b}$ (where a and b are coprime), then the value of $a^2+b^2$ is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

1

Answer 2

We are given

$$\sin^{-1}\sqrt{1-x^2}=\tan^{-1}\frac{\sqrt{2-1}}{x}.$$

Notice that $\sqrt{2-1}=\sqrt{1}=1$. Thus the equation becomes

$$\sin^{-1}\sqrt{1-x^2}=\tan^{-1}\frac{1}{x}\,.$$

Step 1. Rewrite the left‐side:

Recall that for $0\le x\le1$ we have

$$\sin^{-1}\sqrt{1-x^2}=\cos^{-1}x\,.$$

Thus the equation is equivalent to

$$\cos^{-1} x=\tan^{-1}\frac{1}{x}\,.$$

Step 2. Express right‐side in terms of inverse cosine:

We know the complementary angle identity

$$\tan^{-1}\frac{1}{x}=\frac{\pi}{2}-\tan^{-1}x\quad\text{for }x>0,$$

but a more direct way here is to take tangent on both sides of

$$\cos^{-1} x = \tan^{-1}\frac{1}{x}.$$

Let $\theta=\cos^{-1}x$. Then

• $x=\cos\theta$, and
• $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}$.

Now,

$$\tan (\cos^{-1}x)=\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^2}}{x}\,.$$

On the other side, since $\theta=\tan^{-1}(1/x)$ we have

$$\tan\theta=\frac{1}{x}\,.$$

Thus, equate:

$$\frac{\sqrt{1-x^2}}{x}=\frac{1}{x}\,.$$

Step 3. Solve for $x$:

Since $x\neq 0$ appears in the denominator, multiplying both sides by $x$ (and we will later discuss the $x=0$ case separately) gives

$$\sqrt{1-x^2}=1\,.$$

Squaring both sides,

$$1-x^2=1\quad\Longrightarrow\quad x^2=0\,,$$

so

$$x=0\,.$$

Step 4. Check the solution $x=0$:

If we set $x=0$ directly, note that

$$\sin^{-1}\sqrt{1-0^2}=\sin^{-1}(1)=\frac{\pi}{2}\,.$$

On the other hand,

$$\tan^{-1}\frac{1}{0}$$

is not defined in the usual sense but we interpret $\frac{1}{0}$ in the limiting sense as $+\infty$, and

$$\tan^{-1}(+\infty)=\frac{\pi}{2}\,.$$

Thus, $x=0$ is acceptable as the solution in the limiting sense.

Since the problem states “$x=\frac{a}{b}$” with $a$ and $b$ coprime, we express $0$ as $\frac{0}{1}$. Then

$$a^2+b^2=0^2+1^2=1\,.$$