The value of `cot(cosec^(-1)(5)/(3)+tan^(-1)""(2)/(3))` is

Options

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

Answer

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

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

By definition, if \( \theta = \csc^{-1}\left(\frac{5}{3}\right) \), then:
\[ \csc \theta = \frac{5}{3} \]
This implies:
\[ \sin \theta = \frac{3}{5} \]

### Step 2: Find \( \cos \theta \) and \( \tan \theta \)

Using the Pythagorean identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting \( \sin \theta = \frac{3}{5} \):
\[ \cos^2 \theta + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \cos^2 \theta + \frac{9}{25} = 1 \]
\[ \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]
Thus:
\[ \cos \theta = \frac{4}{5} \]

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

### Step 3: Rewrite the expression

Now we can rewrite the original expression:
\[ \cot\left(\csc^{-1}\left(\frac{5}{3}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) = \cot\left(\theta + \tan^{-1}\left(\frac{2}{3}\right)\right) \]

### Step 4: Use the cotangent addition formula

Using the cotangent addition formula:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
Let \( A = \theta \) and \( B = \tan^{-1}\left(\frac{2}{3}\right) \).

We know:
\[ \cot \theta = \frac{1}{\tan \theta} = \frac{4}{3} \]
And for \( B \):
\[ \tan B = \frac{2}{3} \implies \cot B = \frac{3}{2} \]

### Step 5: Substitute into the formula

Now substituting into the cotangent addition formula:
\[ \cot(A + B) = \frac{\frac{4}{3} \cdot \frac{3}{2} - 1}{\frac{4}{3} + \frac{3}{2}} \]

Calculating the numerator:
\[ \frac{4}{3} \cdot \frac{3}{2} = \frac{12}{6} = 2 \]
Thus:
\[ \text{Numerator} = 2 - 1 = 1 \]

Calculating the denominator:
\[ \frac{4}{3} + \frac{3}{2} = \frac{8}{6} + \frac{9}{6} = \frac{17}{6} \]

### Step 6: Final calculation

Now we have:
\[ \cot(A + B) = \frac{1}{\frac{17}{6}} = \frac{6}{17} \]

### Conclusion

Thus, the value of \( \cot\left(\csc^{-1}\left(\frac{5}{3}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) \) is:
\[ \boxed{\frac{6}{17}} \]