If `tan(sin^(-1)sqrt(1-x^2))=sin(tan^(-1)2)` then x is

Options

A. `-sqrt5/3`
B. `3/sqrt10`
C. `sqrt5/3`
D. Both (1) & (2)

Answer

To solve the equation \( \tan(\sin^{-1}(\sqrt{1-x^2})) = \sin(\tan^{-1}(2)) \), we will follow these steps:

### Step 1: Simplify the Left Side
Let \( \theta = \sin^{-1}(\sqrt{1-x^2}) \). Then, we have:
\[ \sin(\theta) = \sqrt{1-x^2} \]
Using the Pythagorean identity, we can find \( \cos(\theta) \):
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = x \]
Now, we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sqrt{1-x^2}}{x} \]

### Step 2: Simplify the Right Side
Now, we simplify the right side \( \sin(\tan^{-1}(2)) \). Let \( \phi = \tan^{-1}(2) \). Then, we have:
\[ \tan(\phi) = 2 \]
Using the definition of tangent, we can form a right triangle where the opposite side is 2 and the adjacent side is 1. The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{2^2 + 1^2} = \sqrt{5} \]
Now, we can find \( \sin(\phi) \):
\[ \sin(\phi) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}} \]

### Step 3: Set the Two Sides Equal
Now we equate both sides:
\[ \frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}} \]

### Step 4: Cross Multiply
Cross multiplying gives:
\[ \sqrt{5} \cdot \sqrt{1-x^2} = 2x \]

### Step 5: Square Both Sides
Squaring both sides results in:
\[ 5(1-x^2) = 4x^2 \]
Expanding this gives:
\[ 5 - 5x^2 = 4x^2 \]

### Step 6: Rearranging the Equation
Combining like terms results in:
\[ 5 = 9x^2 \]
Thus, we can solve for \( x^2 \):
\[ x^2 = \frac{5}{9} \]

### Step 7: Taking the Square Root
Taking the square root of both sides gives:
\[ x = \pm \frac{\sqrt{5}}{3} \]

### Final Answer
Thus, the values of \( x \) are:
\[ x = \frac{\sqrt{5}}{3} \quad \text{or} \quad x = -\frac{\sqrt{5}}{3} \]