`tan(cos^(-1)x)=`

Options

A. `sqrt(1-x^(2))/x`
B. `x/(1+x^(2))`
C. `sqrt(1+x^(2))/x`
D. `sqrt(1-x^(2))`

Answer

To solve the problem \( \tan(\cos^{-1} x) \), we can follow these steps:

### Step 1: Let \( \theta = \cos^{-1} x \)
This means that \( x = \cos \theta \).

### Step 2: Construct a right triangle
In a right triangle, if \( \theta \) is one of the angles, then:
- The adjacent side (base) to angle \( \theta \) is \( x \).
- The hypotenuse is \( 1 \) (since \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)).

### Step 3: Find the length of the opposite side
Using the Pythagorean theorem:
\[ \text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2 \]
Let the length of the opposite side be \( h \). Then:
\[ h^2 + x^2 = 1^2 \]
\[ h^2 = 1 - x^2 \]
\[ h = \sqrt{1 - x^2} \]

### Step 4: Calculate \( \tan \theta \)
The tangent of angle \( \theta \) is given by:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{x} = \frac{\sqrt{1 - x^2}}{x} \]

### Step 5: Substitute back to find \( \tan(\cos^{-1} x) \)
Thus, we have:
\[ \tan(\cos^{-1} x) = \frac{\sqrt{1 - x^2}}{x} \]

### Final Answer
Therefore, the value of \( \tan(\cos^{-1} x) \) is:
\[ \boxed{\frac{\sqrt{1 - x^2}}{x}} \]

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