`tan(sin^(-1)((3)/(5))+cos^(-1)((3)/(sqrt(13)))=`

Options

A. `(sqrt(13))/(5)`
B. `(5)/(sqrt(13))`
C. `(17)/(6)`
D. `(17)/(9)`

Answer

To solve the problem \( \tan(\sin^{-1}(\frac{3}{5}) + \cos^{-1}(\frac{3}{\sqrt{13}})) \), we will follow these steps:

### Step 1: Define the angles
Let:
- \( \alpha = \sin^{-1}(\frac{3}{5}) \)
- \( \beta = \cos^{-1}(\frac{3}{\sqrt{13}}) \)

### Step 2: Find \( \tan(\alpha) \) and \( \tan(\beta) \)
From the definition of sine and cosine:
- For \( \alpha \):
\[ \sin(\alpha) = \frac{3}{5} \implies \text{Opposite} = 3, \text{Hypotenuse} = 5 \]
Using Pythagorean theorem:
\[ \text{Adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Therefore:
\[ \tan(\alpha) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \]

- For \( \beta \):
\[ \cos(\beta) = \frac{3}{\sqrt{13}} \implies \text{Adjacent} = 3, \text{Hypotenuse} = \sqrt{13} \]
Using Pythagorean theorem:
\[ \text{Opposite} = \sqrt{(\sqrt{13})^2 - 3^2} = \sqrt{13 - 9} = \sqrt{4} = 2 \]
Therefore:
\[ \tan(\beta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{3} \]

### Step 3: Use the tangent addition formula
We can now use the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)} \]
Substituting the values:
\[ \tan(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right)} \]

### Step 4: Simplify the numerator
Finding a common denominator for the numerator:
\[ \tan(\alpha + \beta) = \frac{\frac{3 \cdot 3}{12} + \frac{2 \cdot 4}{12}}{1 - \frac{6}{12}} = \frac{\frac{9 + 8}{12}}{1 - \frac{1}{2}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \cdot 2 = \frac{17}{6} \]

### Final Result
Thus, we have:
\[ \tan(\sin^{-1}(\frac{3}{5}) + \cos^{-1}(\frac{3}{\sqrt{13}})) = \frac{17}{6} \]