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

Options

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

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: Simplify \( 2 \tan^{-1}\left(\frac{1}{5}\right) \)

Using the double angle formula for tangent, we have:
\[ 2 \tan^{-1}(x) = \tan^{-1\left(\frac{2x}{1-x^2}\right)} \]
Let \( x = \frac{1}{5} \):
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right) \]
Calculating the denominator:
\[ 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \]
Now substituting back:
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{2 \cdot 25}{5 \cdot 24}\right) = \tan^{-1}\left(\frac{10}{24}\right) = \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 2: Substitute back into the original expression

Now we can rewrite the original expression:
\[ \cos^{-1}\left(\frac{15}{17}\right) + 2 \tan^{-1}\left(\frac{1}{5}\right) = \cos^{-1}\left(\frac{15}{17}\right) + \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 3: Let \( \cos^{-1}\left(\frac{15}{17}\right) = \theta \)

This means:
\[ \cos(\theta) = \frac{15}{17} \]
Using the Pythagorean theorem in a right triangle where the adjacent side is 15 and the hypotenuse is 17, we find the opposite side:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \implies \sin^2(\theta) = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289} \]
Thus,
\[ \sin(\theta) = \frac{8}{17} \]

### Step 4: Find \( \tan(\theta) \)

Now we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{8}{17}}{\frac{15}{17}} = \frac{8}{15} \]
So we have:
\[ \theta = \tan^{-1}\left(\frac{8}{15}\right) \]

### Step 5: Combine the two \( \tan^{-1} \) terms

Now we have:
\[ \tan^{-1}\left(\frac{8}{15}\right) + \tan^{-1}\left(\frac{5}{12}\right) \]
Using the formula for the sum of two arctangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Let \( x = \frac{8}{15} \) and \( y = \frac{5}{12} \):
\[ \frac{x+y}{1-xy} = \frac{\frac{8}{15} + \frac{5}{12}}{1 - \left(\frac{8}{15} \cdot \frac{5}{12}\right)} \]
Calculating the numerator:
\[ \frac{8 \cdot 12 + 5 \cdot 15}{15 \cdot 12} = \frac{96 + 75}{180} = \frac{171}{180} \]
Calculating the denominator:
\[ 1 - \frac{40}{180} = \frac{140}{180} = \frac{7}{9} \]
Thus:
\[ \tan^{-1}\left(\frac{171/180}{7/9}\right) = \tan^{-1}\left(\frac{171 \cdot 9}{140 \cdot 10}\right) = \tan^{-1}\left(\frac{171}{140}\right) \]

### Step 6: Final Result

Thus, we have:
\[ \cos^{-1}\left(\frac{15}{17}\right) + 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{171}{140}\right) \]