If `tan(sec^(-1)x)=sin("cos"^(-1)1/sqrt5)`, then x is equal to:

Options

A. `+-3/sqrt5`
B. `+-sqrt5/3`
C. `+-sqrt3/3`
D. None of these

Answer

To solve the equation \( \tan(\sec^{-1} x) = \sin(\cos^{-1} \frac{1}{\sqrt{5}}) \), we will follow these steps:

### Step 1: Understand the left-hand side
We start with the left-hand side, \( \tan(\sec^{-1} x) \).

Using the identity:
\[ \tan(\sec^{-1} x) = \sqrt{x^2 - 1} \]
This is derived from the right triangle definition where \( \sec \theta = x \) implies that the adjacent side is 1 and the hypotenuse is \( x \). Therefore, the opposite side can be calculated using the Pythagorean theorem:
\[ \text{Opposite} = \sqrt{x^2 - 1} \]
Thus, we have:
\[ \tan(\sec^{-1} x) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{x^2 - 1}}{1} = \sqrt{x^2 - 1} \]

### Step 2: Understand the right-hand side
Now, we analyze the right-hand side, \( \sin(\cos^{-1} \frac{1}{\sqrt{5}}) \).

Using the identity:
\[ \sin(\cos^{-1} y) = \sqrt{1 - y^2} \]
Substituting \( y = \frac{1}{\sqrt{5}} \):
\[ \sin(\cos^{-1} \frac{1}{\sqrt{5}}) = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

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

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ x^2 - 1 = \left(\frac{2}{\sqrt{5}}\right)^2 \]
Calculating the right-hand side:
\[ x^2 - 1 = \frac{4}{5} \]

### Step 5: Solve for \( x^2 \)
Now, we add 1 to both sides:
\[ x^2 = \frac{4}{5} + 1 = \frac{4}{5} + \frac{5}{5} = \frac{9}{5} \]

### Step 6: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm \sqrt{\frac{9}{5}} = \pm \frac{3}{\sqrt{5}} \]

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