`tan(cos^(-1)((4)/(5))+tan^(-1)((2)/(3)))=`

Options

A. `(17)/(24)`
B. `(24)/(17)`
C. `(17)/(6)`
D. `(6)/(17)`

Answer

To solve the problem \( \tan(\cos^{-1}(\frac{4}{5}) + \tan^{-1}(\frac{2}{3})) \), we will follow these steps:

### Step 1: Let \( \alpha = \cos^{-1}(\frac{4}{5}) \) and \( \beta = \tan^{-1}(\frac{2}{3}) \)

We start by defining \( \alpha \) and \( \beta \) for clarity.

### Step 2: Find \( \tan(\alpha) \)

Using the definition of cosine, we know:

\[ \cos(\alpha) = \frac{4}{5} \]

To find \( \tan(\alpha) \), we need to find the opposite side of the triangle. We can use the Pythagorean theorem:

\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \]

Let \( \sin(\alpha) = y \). Then:

\[ y^2 + \left(\frac{4}{5}\right)^2 = 1 \]

Calculating \( \left(\frac{4}{5}\right)^2 \):

\[ y^2 + \frac{16}{25} = 1 \implies y^2 = 1 - \frac{16}{25} = \frac{9}{25} \implies y = \frac{3}{5} \]

Thus,

\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 3: Find \( \tan(\beta) \)

From the definition of tangent, we have:

\[ \tan(\beta) = \frac{2}{3} \]

### Step 4: Use the tangent addition formula

Now, we can use the tangent addition formula:

\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)} \]

Substituting the values we found:

\[ \tan(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right)} \]

### Step 5: Simplify the numerator

To add \( \frac{3}{4} \) and \( \frac{2}{3} \), we need a common denominator:

\[ \text{Common denominator} = 12 \]

Thus,

\[ \frac{3}{4} = \frac{9}{12}, \quad \frac{2}{3} = \frac{8}{12} \]

So,

\[ \tan(\alpha + \beta) = \frac{\frac{9}{12} + \frac{8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \cdot 2 = \frac{17}{6} \]

### Final Answer

Thus,

\[ \tan(\cos^{-1}(\frac{4}{5}) + \tan^{-1}(\frac{2}{3})) = \frac{17}{6} \]