`sin^(-1)" 1/sqrt5 +cot^(-1)3` is equal to

Options

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

Answer

To solve the expression \( \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) + \cot^{-1}(3) \), we can follow these steps:

### Step 1: Convert \( \cot^{-1}(3) \) to a form involving \( \tan \)
Using the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \), we can rewrite:
\[ \cot^{-1}(3) = \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Let \( A = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \)
This implies:
\[ \sin A = \frac{1}{\sqrt{5}} \]
To find \( \cos A \), we can use the Pythagorean identity:
\[ \cos^2 A = 1 - \sin^2 A \]
Calculating \( \sin^2 A \):
\[ \sin^2 A = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \]
Thus,
\[ \cos^2 A = 1 - \frac{1}{5} = \frac{4}{5} \]
So, we have:
\[ \cos A = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

### Step 3: Find \( \tan A \)
Using the definition of tangent:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]

### Step 4: Combine \( A \) and \( \tan^{-1}\left(\frac{1}{3}\right) \)
Now we can express the sum:
\[ \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \tan^{-1}\left(\frac{1}{3}\right) \]
Using the formula for the sum of two angles in tangent:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting \( \tan A = \frac{1}{2} \) and \( \tan B = \frac{1}{3} \):
\[ \tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right)} \]

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

### Step 6: Find the angle
Since \( \tan(A + B) = 1 \), we have:
\[ A + B = \tan^{-1}(1) = \frac{\pi}{4} \]

### Conclusion
Therefore, the value of \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \cot^{-1}(3) \) is:
\[ \frac{\pi}{4} \]