The value of x where `xgt0 and tan(sec^(-1)(1/x))=sin(tan^(-1)2)` is

Options

A. `sqrt(5)`
B. `sqrt(5)/(3)`
C. 1
D. `2/3`

Explanation

To solve the equation \( \tan(\sec^{-1}(1/x)) = \sin(\tan^{-1}(2)) \) for \( x > 0 \), we can follow these steps:

### Step 1: Set up the equation
Let \( \theta = \sec^{-1}(1/x) \). Then we have:
\[ \sec(\theta) = \frac{1}{x} \]

### Step 2: Find the relationship between \( \tan(\theta) \) and \( x \)
Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we can express \( \cos(\theta) \) as:
\[ \cos(\theta) = x \]
Using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we can find \( \tan(\theta) \):
\[ \tan^2(\theta) = \sec^2(\theta) - 1 = \left(\frac{1}{x}\right)^2 - 1 = \frac{1 - x^2}{x^2} \]
Thus,
\[ \tan(\theta) = \frac{\sqrt{1 - x^2}}{x} \]

### Step 3: Evaluate the right-hand side
Now, we need to evaluate \( \sin(\tan^{-1}(2)) \). Let \( \alpha = \tan^{-1}(2) \), then:
\[ \tan(\alpha) = 2 \]
This means we can construct a right triangle where the opposite side is 2 and the adjacent side is 1. The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
Now, we can find \( \sin(\alpha) \):
\[ \sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}} \]

### Step 4: Set the two sides equal
Now we have:
\[ \tan(\sec^{-1}(1/x)) = \sin(\tan^{-1}(2) \]
This gives us:
\[ \frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}} \]

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

### Step 6: Square both sides
Squaring both sides results in:
\[ 5(1 - x^2) = 4x^2 \]
Expanding this gives:
\[ 5 - 5x^2 = 4x^2 \]

### Step 7: Rearrange the equation
Rearranging the equation results in:
\[ 5 = 9x^2 \]
Thus,
\[ x^2 = \frac{5}{9} \]

### Step 8: Solve for \( x \)
Taking the square root gives:
\[ x = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Final Answer
The value of \( x \) is:
\[ \boxed{\frac{\sqrt{5}}{3}} \]