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

Options

A. `tan^(-1)((67)/(19))`
B. `tan^(-1)((1)/(59))`
C. `tan^(-1)((19)/(67))`
D. `tan^(-1)(59)`

Answer

To solve the equation \( \tan^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \), we can follow these steps:

### Step 1: Let \( y = \tan^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \)

### Step 2: Convert \( \cos^{-1}\left(\frac{3}{5}\right) \) into \( \tan^{-1} \)

Let \( \theta = \cos^{-1}\left(\frac{3}{5}\right) \). Then, by definition of cosine:
\[ \cos \theta = \frac{3}{5} \]

### Step 3: Use the Pythagorean theorem to find \( \sin \theta \)

In a right triangle, if the adjacent side is 3 and the hypotenuse is 5, we can find the opposite side using:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
\[ \sin^2 \theta + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{9}{25} = 1 \]
\[ \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin \theta = \frac{4}{5} \]

### Step 4: Find \( \tan \theta \)

Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 5: Substitute back into the equation

Now we can rewrite \( y \):
\[ y = \tan^{-1}\left(\frac{5}{13}\right) + \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Use the formula for the sum of arctangents

Using the formula:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
where \( a = \frac{5}{13} \) and \( b = \frac{4}{3} \):
\[ y = \tan^{-1}\left(\frac{\frac{5}{13} + \frac{4}{3}}{1 - \frac{5}{13} \cdot \frac{4}{3}}\right) \]

### Step 7: Calculate the numerator

Finding a common denominator for the numerator:
\[ \frac{5}{13} + \frac{4}{3} = \frac{5 \cdot 3 + 4 \cdot 13}{39} = \frac{15 + 52}{39} = \frac{67}{39} \]

### Step 8: Calculate the denominator

Calculating the denominator:
\[ 1 - \frac{5}{13} \cdot \frac{4}{3} = 1 - \frac{20}{39} = \frac{39 - 20}{39} = \frac{19}{39} \]

### Step 9: Substitute back into the formula

Now substituting back:
\[ y = \tan^{-1}\left(\frac{\frac{67}{39}}{\frac{19}{39}}\right) = \tan^{-1}\left(\frac{67}{19}\right) \]

### Final Answer

Thus, we have:
\[ y = \tan^{-1}\left(\frac{67}{19}\right) \]

### Summary of the Answer
The final answer is:
\[ \tan^{-1}\left(\frac{67}{19}\right) \]