`cot(cosec ^(-1)(5/3) + tan^(-1)(2/3))=`

Options

A. `3/17`
B. `4/17`
C. `5/17`
D. `6/17`

Answer

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

### Step 1: Let \( \theta = \csc^{-1}\left(\frac{5}{3}\right) \)
This means that \( \csc(\theta) = \frac{5}{3} \).

### Step 2: Convert \( \csc(\theta) \) to \( \sin(\theta) \)
From the definition of cosecant, we have:
\[ \sin(\theta) = \frac{1}{\csc(\theta)} = \frac{3}{5} \]

### Step 3: Use the Pythagorean theorem to find \( \cos(\theta) \)
We can form a right triangle where the opposite side is 3 and the hypotenuse is 5. To find the adjacent side (base), we use:
\[ \text{base} = \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Thus, \( \cos(\theta) = \frac{4}{5} \).

### Step 4: Find \( \tan(\theta) \)
Using the definition of tangent:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \]

### Step 5: Substitute \( \tan(\theta) \) into the expression
Now we can express \( \cot\left(\csc^{-1}\left(\frac{5}{3}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) \) as:
\[ \cot\left(\theta + \tan^{-1}\left(\frac{2}{3}\right)\right) \]

### Step 6: Use the formula for \( \tan^{-1}(a) + \tan^{-1}(b) \)
Using the formula:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
where \( a = \tan(\theta) = \frac{3}{4} \) and \( b = \frac{2}{3} \).

### Step 7: Calculate \( a + b \) and \( 1 - ab \)
1. Calculate \( a + b \):
\[ \frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12} \]

2. Calculate \( ab \):
\[ ab = \frac{3}{4} \cdot \frac{2}{3} = \frac{6}{12} = \frac{1}{2} \]
Thus, \( 1 - ab = 1 - \frac{1}{2} = \frac{1}{2} \).

### Step 8: Substitute into the formula
Now we have:
\[ \tan^{-1}\left(\frac{\frac{17}{12}}{\frac{1}{2}}\right) = \tan^{-1}\left(\frac{17 \cdot 2}{12}\right) = \tan^{-1}\left(\frac{34}{12}\right) = \tan^{-1}\left(\frac{17}{6}\right) \]

### Step 9: Find \( \cot\left(\theta + \tan^{-1}\left(\frac{2}{3}\right)\right) \)
Thus:
\[ \cot\left(\theta + \tan^{-1}\left(\frac{2}{3}\right)\right) = \cot\left(\tan^{-1}\left(\frac{17}{6}\right)\right) \]
Using the identity \( \cot(\tan^{-1}(x)) = \frac{1}{x} \):
\[ \cot\left(\tan^{-1}\left(\frac{17}{6}\right)\right) = \frac{6}{17} \]

### Final Answer
Therefore, the final answer is:
\[ \cot\left(\csc^{-1}\left(\frac{5}{3}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) = \frac{6}{17} \]