If A is in the fourth quadrant and `cos A=(5)/(13)`, find the value of `(13 sin A+5 secA)/(5tanA+6" cosec"A)`.

Answer

To solve the problem, we need to find the value of the expression \((13 \sin A + 5 \sec A) / (5 \tan A + 6 \csc A)\) given that \(\cos A = \frac{5}{13}\) and \(A\) is in the fourth quadrant.

### Step 1: Find \(\sin A\)

Using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]
We can substitute \(\cos A\):
\[ \sin^2 A + \left(\frac{5}{13}\right)^2 = 1 \]
\[ \sin^2 A + \frac{25}{169} = 1 \]
\[ \sin^2 A = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \]
Taking the square root:
\[ \sin A = \pm \frac{12}{13} \]
Since \(A\) is in the fourth quadrant, \(\sin A\) is negative:
\[ \sin A = -\frac{12}{13} \]

### Step 2: Find \(\sec A\)

The secant function is the reciprocal of the cosine function:
\[ \sec A = \frac{1}{\cos A} = \frac{1}{\frac{5}{13}} = \frac{13}{5} \]

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

The tangent function is defined as:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5} \]

### Step 4: Find \(\csc A\)

The cosecant function is the reciprocal of the sine function:
\[ \csc A = \frac{1}{\sin A} = \frac{1}{-\frac{12}{13}} = -\frac{13}{12} \]

### Step 5: Substitute values into the expression

Now we substitute the values into the expression:
\[ \frac{13 \sin A + 5 \sec A}{5 \tan A + 6 \csc A} \]
Substituting the values we found:
\[ = \frac{13 \left(-\frac{12}{13}\right) + 5 \left(\frac{13}{5}\right)}{5 \left(-\frac{12}{5}\right) + 6 \left(-\frac{13}{12}\right)} \]

### Step 6: Simplify the numerator

Calculating the numerator:
\[ = \frac{-12 + 13}{-12 - \frac{78}{12}} \]
\[ = \frac{1}{-12 - 6.5} = \frac{1}{-18.5} \]

### Step 7: Simplify the denominator

Calculating the denominator:
\[ = -12 - 6.5 = -18.5 \]

### Step 8: Final calculation

Now we have:
\[ = \frac{1}{-18.5} = -\frac{2}{37} \]

Thus, the final answer is:
\[ \boxed{-\frac{2}{37}} \]