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) & (3)

Answer

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

### Step 1: Define the variables
Let \( \theta = \sin^{-1}(\sqrt{1-x^2}) \). This implies that \( \sin(\theta) = \sqrt{1-x^2} \).

**Hint:** Use the definition of the sine function to relate it to a right triangle.

### Step 2: Construct a right triangle
In a right triangle, if \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), we can set:
- Opposite side = \( \sqrt{1-x^2} \)
- Hypotenuse = 1

Now, we need to find the base of the triangle using the Pythagorean theorem:
\[ \text{Base} = \sqrt{1^2 - (\sqrt{1-x^2})^2} = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = x. \]

**Hint:** Remember the Pythagorean theorem: \( a^2 + b^2 = c^2 \).

### Step 3: Find \( \tan(\theta) \)
Now we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1-x^2}}{x}. \]

**Hint:** Recall the definition of tangent in terms of sine and cosine.

### Step 4: Define the second variable
Let \( \alpha = \tan^{-1}(2) \). This means that \( \tan(\alpha) = 2 \).

**Hint:** Use the definition of tangent to set up a triangle for \( \alpha \).

### Step 5: Construct a right triangle for \( \alpha \)
In a right triangle:
- Opposite side = 2
- Adjacent side = 1

Now, we can find the hypotenuse:
\[ \text{Hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}. \]

**Hint:** Again, apply the Pythagorean theorem.

### Step 6: Find \( \sin(\alpha) \)
Now, we find \( \sin(\alpha) \):
\[ \sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}. \]

**Hint:** Use the sine definition based on the triangle you just constructed.

### Step 7: Set up the equation
Now, we have:
\[ \tan(\theta) = \sin(\alpha). \]
Substituting the values we found:
\[ \frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}}. \]

**Hint:** This equation relates the two expressions we derived.

### Step 8: Cross-multiply
Cross-multiplying gives:
\[ \sqrt{5} \cdot \sqrt{1-x^2} = 2x. \]

**Hint:** Be careful with the algebra when cross-multiplying.

### Step 9: Square both sides
Squaring both sides results in:
\[ 5(1-x^2) = 4x^2. \]

**Hint:** Distributing the 5 will help simplify the equation.

### Step 10: Rearrange the equation
Rearranging gives:
\[ 5 - 5x^2 = 4x^2 \implies 5 = 9x^2 \implies x^2 = \frac{5}{9}. \]

**Hint:** Isolate \( x^2 \) to solve for \( x \).

### Step 11: Solve for \( x \)
Taking the square root gives:
\[ x = \pm \frac{\sqrt{5}}{3}. \]

**Hint:** Remember to consider both the positive and negative roots.

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