If the numerical value of `tan(cos^(-1)((4)/(5))+tan^(-1)((2)/(3)))" is " ((a)/(b))`, where a, b are two positive integers and their H.C.F. is 1

Options

A. `a+b=23`
B. `a-b=11`
C. `3b=a+1`
D. `2a=3b`

Answer

To solve the problem, we need to find the numerical value of \( \tan(\cos^{-1}(\frac{4}{5}) + \tan^{-1}(\frac{2}{3})) \) and express it in the form \( \frac{a}{b} \) where \( a \) and \( b \) are positive integers with their H.C.F. being 1.

### Step-by-Step Solution:

1. **Identify the angles**:
Let \( \theta = \cos^{-1}(\frac{4}{5}) \) and \( \phi = \tan^{-1}(\frac{2}{3}) \). We need to find \( \tan(\theta + \phi) \).

2. **Use the right triangle for \( \theta \)**:
From \( \theta = \cos^{-1}(\frac{4}{5}) \), we can visualize a right triangle where:
- Adjacent side (base) = 4
- Hypotenuse = 5
- Using Pythagoras theorem, the opposite side (perpendicular) can be calculated as:
\[ \text{Opposite} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Thus, \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \).

3. **Calculate \( \tan(\phi) \)**:
From \( \phi = \tan^{-1}(\frac{2}{3}) \), we have:
\[ \tan(\phi) = \frac{2}{3} \]

4. **Use the tangent addition formula**:
The formula for \( \tan(\theta + \phi) \) is:
\[ \tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta) \tan(\phi)} \]
Substituting the values:
\[ \tan(\theta + \phi) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4} \cdot \frac{2}{3}\right)} \]

5. **Calculate the numerator**:
To add \( \frac{3}{4} \) and \( \frac{2}{3} \), we find a common denominator:
\[ \frac{3}{4} = \frac{9}{12}, \quad \frac{2}{3} = \frac{8}{12} \]
Thus,
\[ \tan(\theta + \phi) = \frac{\frac{9}{12} + \frac{8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \cdot 2 = \frac{17}{6} \]

6. **Identify \( a \) and \( b \)**:
From \( \tan(\theta + \phi) = \frac{17}{6} \), we have \( a = 17 \) and \( b = 6 \).

7. **Check the H.C.F.**:
The H.C.F. of 17 and 6 is 1, confirming they are co-prime.

### Final Answer:
The numerical value of \( \tan(\cos^{-1}(\frac{4}{5}) + \tan^{-1}(\frac{2}{3})) \) is \( \frac{17}{6} \).