Find the value of
`tan[cos^(-1) (1/2) + tan^(-1) ( - 1/sqrt3)]`

Options

A. `-1/sqrt2 `
B. `-1/sqrt3`
C. `1/sqrt3`
D. `1/sqrt2`

Answer

To find the value of \( \tan\left(\cos^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right) \), we can follow these steps:

### Step 1: Identify \( \theta \) for \( \cos^{-1}\left(\frac{1}{2}\right) \)
Let \( \theta = \cos^{-1}\left(\frac{1}{2}\right) \). This means that \( \cos(\theta) = \frac{1}{2} \).

### Step 2: Determine the values of sine and tangent
From the definition of cosine, we can construct a right triangle where:
- The adjacent side (base) is \( 1 \)
- The hypotenuse is \( 2 \)

Using the Pythagorean theorem, the opposite side (perpendicular) can be calculated as:
\[ \text{opposite} = \sqrt{(2^2 - 1^2)} = \sqrt{3} \]
Thus, we have:
- \( \sin(\theta) = \frac{\sqrt{3}}{2} \)
- \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3} \)

### Step 3: Identify \( \phi \) for \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
Let \( \phi = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \). This means that:
\[ \tan(\phi) = -\frac{1}{\sqrt{3}} \]
The angle whose tangent is \( -\frac{1}{\sqrt{3}} \) corresponds to \( -30^\circ \) or \( -\frac{\pi}{6} \).

### Step 4: Use the tangent addition formula
We need to find:
\[ \tan(\theta + \phi) \]
Using the tangent addition formula:
\[ \tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)} \]
Substituting the values we found:
\[ \tan(\theta + \phi) = \frac{\sqrt{3} + \left(-\frac{1}{\sqrt{3}}\right)}{1 - \left(\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)} \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]
Calculating the denominator:
\[ 1 + 1 = 2 \]
Thus, we have:
\[ \tan(\theta + \phi) = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}} \]

### Final Answer
Therefore, the value of \( \tan\left(\cos^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right) \) is:
\[ \frac{1}{\sqrt{3}} \]

---