The sum of the two angles `cot^(-1) 3 and cosec^(-1) sqrt(5)` is

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Answer

To solve the problem of finding the sum of the two angles \( \cot^{-1}(3) \) and \( \csc^{-1}(\sqrt{5}) \), we will follow these steps:

### Step 1: Define the angles
Let:
- \( A = \cot^{-1}(3) \)
- \( B = \csc^{-1}(\sqrt{5}) \)

### Step 2: Convert \( A \) to \( \tan^{-1} \)
Using the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \):
\[ A = \cot^{-1}(3) = \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 3: Convert \( B \) to a right triangle
Since \( B = \csc^{-1}(\sqrt{5}) \), we know that:
\[ \csc(B) = \sqrt{5} \implies \sin(B) = \frac{1}{\sqrt{5}} \]
Using the Pythagorean theorem, we can find the adjacent side:
- Hypotenuse = \( \sqrt{5} \)
- Opposite side = 1 (since \( \sin(B) = \frac{1}{\sqrt{5}} \))

Now, using the Pythagorean theorem:
\[ \text{Adjacent}^2 + 1^2 = (\sqrt{5})^2 \implies \text{Adjacent}^2 + 1 = 5 \implies \text{Adjacent}^2 = 4 \implies \text{Adjacent} = 2 \]

### Step 4: Find \( \tan(B) \)
Now we can find \( \tan(B) \):
\[ \tan(B) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{2} \]
Thus, we can write:
\[ B = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 5: Sum of angles
Now we need to find \( A + B \):
\[ A + B = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 6: Use the tangent addition formula
Using the formula for the sum of two arctangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
where \( x = \frac{1}{3} \) and \( y = \frac{1}{2} \):
\[ A + B = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} \cdot \frac{1}{2}}\right) \]

### Step 7: Simplify the expression
Calculating the numerator:
\[ \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]
Calculating the denominator:
\[ 1 - \frac{1}{3} \cdot \frac{1}{2} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus,
\[ A + B = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

### Step 8: Final result
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we conclude that:
\[ A + B = \frac{\pi}{4} \]

### Summary
The sum of the angles \( \cot^{-1}(3) \) and \( \csc^{-1}(\sqrt{5}) \) is:
\[ \frac{\pi}{4} \]