Value of x satisfying ` tan(sec^(- 1)x)=sin(cos^(- 1)(1/(sqrt(5))))`

Options

A. `x=-3/sqrt5`
B. `x=sqrt5/2`
C. `x=3/sqrt5`
D. `x=-sqrt5/2`

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 side of the equation
Let \( \theta = \sec^{-1} x \). By the definition of secant, we have:
\[ \sec \theta = x \implies \cos \theta = \frac{1}{x} \]

### Step 2: Find the value of \( \tan \theta \)
Using the Pythagorean identity, we can find the opposite side of the triangle. In a right triangle where the adjacent side is \( 1 \) (base) and the hypotenuse is \( x \):
\[ \text{Opposite side} = \sqrt{x^2 - 1} \]
Thus, we can express \( \tan \theta \) as:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{x^2 - 1}}{1} = \sqrt{x^2 - 1} \]

### Step 3: Understand the right side of the equation
Now, we need to evaluate \( \sin(\cos^{-1}(\frac{1}{\sqrt{5}})) \). Let \( \phi = \cos^{-1}(\frac{1}{\sqrt{5}}) \). By the definition of cosine:
\[ \cos \phi = \frac{1}{\sqrt{5}} \implies \text{Adjacent} = 1 \quad \text{and} \quad \text{Hypotenuse} = \sqrt{5} \]
Using the Pythagorean theorem, we find the opposite side:
\[ \text{Opposite} = \sqrt{(\sqrt{5})^2 - 1^2} = \sqrt{5 - 1} = \sqrt{4} = 2 \]
Thus, we can express \( \sin \phi \) as:
\[ \sin \phi = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2}{\sqrt{5}} \]

### Step 4: Set the two sides equal
Now we have:
\[ \sqrt{x^2 - 1} = \frac{2}{\sqrt{5}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ x^2 - 1 = \left(\frac{2}{\sqrt{5}}\right)^2 \]
\[ x^2 - 1 = \frac{4}{5} \]

### Step 6: Solve for \( x^2 \)
Adding \( 1 \) to both sides:
\[ x^2 = 1 + \frac{4}{5} = \frac{5}{5} + \frac{4}{5} = \frac{9}{5} \]

### Step 7: Solve for \( x \)
Taking the square root of both sides:
\[ x = \pm \sqrt{\frac{9}{5}} = \pm \frac{3}{\sqrt{5}} \]

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

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