`cos(tan^(-1)((1)/(3))+tan^(-1)((1)/(2)))=`

Options

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

Answer

To solve the problem \( \cos\left(\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right)\right) \), we can use the formula for the sum of two inverse tangents. Here are the steps:

### Step 1: Define the variables
Let:
- \( a = \tan^{-1}\left(\frac{1}{3}\right) \)
- \( b = \tan^{-1}\left(\frac{1}{2}\right) \)

### Step 2: Use the formula for the sum of inverse tangents
The formula for the sum of two inverse tangents is:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]
provided that \( xy < 1 \).

### Step 3: Check the condition
Calculate \( ab \):
\[ ab = \left(\frac{1}{3}\right) \left(\frac{1}{2}\right) = \frac{1}{6} < 1 \]
Since \( ab < 1 \), we can apply the formula.

### Step 4: Apply the formula
Now we can substitute \( x = \frac{1}{3} \) and \( y = \frac{1}{2} \):
\[ \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{6}}\right) \]

### Step 5: Simplify the expression
Calculate the numerator:
\[ \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6} \]
Calculate the denominator:
\[ 1 - \frac{1}{6} = \frac{6 - 1}{6} = \frac{5}{6} \]
Thus, we have:
\[ \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

### Step 6: Find the cosine
Now we know:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
So we need to find:
\[ \cos\left(\tan^{-1}(1)\right) = \cos\left(\frac{\pi}{4}\right) \]

### Step 7: Calculate the cosine
We know:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the final answer is:
\[ \cos\left(\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right)\right) = \frac{1}{\sqrt{2}} \]