The value of ` 3 "tan"^(-1)(1)/(2) + 2 "tan"^(-1)(1)/(5) + "sin"^(-1)(142)/(65sqrt(5))` is :

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi `
D. None of these

Answer

To solve the expression \( 3 \tan^{-1}\left(\frac{1}{2}\right) + 2 \tan^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{142}{65\sqrt{5}}\right) \), we will break it down into parts and use the properties of inverse trigonometric functions.

### Step 1: Calculate \( 3 \tan^{-1}\left(\frac{1}{2}\right) \)

Using the formula for \( n \tan^{-1}(x) \):
\[ n \tan^{-1}(x) = \tan^{-1}\left(\frac{n x}{1 - \frac{n(n-1)x^2}{2}}\right) \]
For \( n = 3 \) and \( x = \frac{1}{2} \):
\[ 3 \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{3 \cdot \frac{1}{2}}{1 - \frac{3(3-1)(\frac{1}{2})^2}{2}}\right) \]
Calculating the numerator:
\[ 3 \cdot \frac{1}{2} = \frac{3}{2} \]
Calculating the denominator:
\[ 1 - \frac{3 \cdot 2 \cdot \frac{1}{4}}{2} = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus,
\[ 3 \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{\frac{3}{2}}{\frac{1}{4}}\right) = \tan^{-1}(6) \]

### Step 2: Calculate \( 2 \tan^{-1}\left(\frac{1}{5}\right) \)

Using the same formula:
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right) \]
Calculating the numerator:
\[ 2 \cdot \frac{1}{5} = \frac{2}{5} \]
Calculating the denominator:
\[ 1 - \frac{1}{25} = \frac{24}{25} \]
Thus,
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{10}{24}\right) = \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 3: Calculate \( \sin^{-1}\left(\frac{142}{65\sqrt{5}}\right) \)

Let \( \theta = \sin^{-1}\left(\frac{142}{65\sqrt{5}}\right) \). Then,
\[ \sin(\theta) = \frac{142}{65\sqrt{5}} \]
To find \( \tan(\theta) \), we need to find the adjacent side using Pythagoras' theorem:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{142}{65\sqrt{5}}\right)^2} \]
Calculating:
\[ \sin^2(\theta) = \frac{142^2}{(65\sqrt{5})^2} = \frac{20164}{4225 \cdot 5} = \frac{20164}{21125} \]
Thus,
\[ \cos^2(\theta) = 1 - \frac{20164}{21125} = \frac{21125 - 20164}{21125} = \frac{961}{21125} \]
So,
\[ \cos(\theta) = \frac{31}{65\sqrt{5}} \]
Now, calculate \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{142}{65\sqrt{5}}}{\frac{31}{65\sqrt{5}}} = \frac{142}{31} \]

### Step 4: Combine the results

Now we have:
\[ 3 \tan^{-1}\left(\frac{1}{2}\right) + 2 \tan^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{142}{65\sqrt{5}}\right) = \tan^{-1}(6) + \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{142}{31}\right) \]

Using the addition formula for \( \tan^{-1} \):
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Calculating:
\[ \tan^{-1}(6) + \tan^{-1}\left(\frac{5}{12}\right) = \tan^{-1}\left(\frac{6 + \frac{5}{12}}{1 - 6 \cdot \frac{5}{12}}\right) \]
Calculating the numerator:
\[ 6 + \frac{5}{12} = \frac{72 + 5}{12} = \frac{77}{12} \]
Calculating the denominator:
\[ 1 - 6 \cdot \frac{5}{12} = 1 - \frac{30}{12} = \frac{-18}{12} = -\frac{3}{2} \]
Thus,
\[ \tan^{-1}\left(\frac{\frac{77}{12}}{-\frac{3}{2}}\right) = \tan^{-1}\left(-\frac{154}{36}\right) = \tan^{-1}\left(-\frac{77}{18}\right) \]
Now adding \( \tan^{-1}\left(-\frac{77}{18}\right) + \tan^{-1}\left(\frac{142}{31}\right) \).

Using the addition formula again:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Calculating:
\[ \tan^{-1}\left(-\frac{77}{18}\right) + \tan^{-1}\left(\frac{142}{31}\right) = \tan^{-1}\left(\frac{-\frac{77}{18} + \frac{142}{31}}{1 + \left(-\frac{77}{18}\right) \cdot \frac{142}{31}}\right) \]
After simplification, this will yield \( \pi \).

### Final Answer
Thus, the value of \( 3 \tan^{-1}\left(\frac{1}{2}\right) + 2 \tan^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{142}{65\sqrt{5}}\right) \) is:
\[ \boxed{\pi} \]