`cos^(-1) ""15/17 + 2 tan ^(-1)""1/5=`

Options

A. `pi/2`
B. `cos^(-1) 171/221`
C. `pi/4`
D. none

Answer

To solve the expression \( \cos^{-1} \left( \frac{15}{17} \right) + 2 \tan^{-1} \left( \frac{1}{5} \right) \), we will follow these steps:

### Step 1: Rewrite \( 2 \tan^{-1} \left( \frac{1}{5} \right) \) using the double angle formula
We know that:
\[ 2 \tan^{-1}(x) = \tan^{-1\left(\frac{2x}{1-x^2}\right)} \]
Let \( x = \frac{1}{5} \). Then:
\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} \right) \]
Calculating the numerator:
\[ 2 \cdot \frac{1}{5} = \frac{2}{5} \]
Calculating the denominator:
\[ 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \]
Thus:
\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \left( \frac{2 \cdot 25}{5 \cdot 24} \right) = \tan^{-1} \left( \frac{10}{24} \right) = \tan^{-1} \left( \frac{5}{12} \right) \]

### Step 2: Combine the angles
Now we have:
\[ \cos^{-1} \left( \frac{15}{17} \right) + \tan^{-1} \left( \frac{5}{12} \right) \]
We can use the identity:
\[ \cos^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{y}{\sqrt{1-x^2}} \right) \]
where \( x = \frac{15}{17} \) and \( y = \frac{5}{12} \).

### Step 3: Calculate \( \sqrt{1 - x^2} \)
First, we need to calculate \( 1 - x^2 \):
\[ 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289} \]
Thus:
\[ \sqrt{1 - x^2} = \sqrt{\frac{64}{289}} = \frac{8}{17} \]

### Step 4: Substitute into the formula
Now we substitute into the identity:
\[ \cos^{-1} \left( \frac{15}{17} \right) + \tan^{-1} \left( \frac{5}{12} \right) = \tan^{-1} \left( \frac{\frac{5}{12}}{\frac{8}{17}} \right) \]
Calculating the right-hand side:
\[ = \tan^{-1} \left( \frac{5 \cdot 17}{12 \cdot 8} \right) = \tan^{-1} \left( \frac{85}{96} \right) \]

### Step 5: Final expression
Thus, we have:
\[ \cos^{-1} \left( \frac{15}{17} \right) + 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{85}{96} \right) \]

### Conclusion
The final answer is:
\[ \tan^{-1} \left( \frac{85}{96} \right) \]