Solution of equation `cot^(-1) x + sin^(-1) . 1/sqrt5 = pi/4 ` is

Options

A. ` x = 3`
B. ` x = 1//sqrt5`
C. ` x = 0`
D. None of these

Answer

To solve the equation \( \cot^{-1} x + \sin^{-1} \frac{1}{\sqrt{5}} = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Rewrite \( \cot^{-1} x \)
Let \( y = \cot^{-1} x \). Then, we can express \( x \) in terms of \( y \):
\[ \cot y = x \implies \tan y = \frac{1}{x} \]
Thus, we can rewrite the equation as:
\[ y + \sin^{-1} \frac{1}{\sqrt{5}} = \frac{\pi}{4} \]

### Step 2: Isolate \( y \)
Rearranging the equation gives:
\[ y = \frac{\pi}{4} - \sin^{-1} \frac{1}{\sqrt{5}} \]

### Step 3: Rewrite \( \sin^{-1} \frac{1}{\sqrt{5}} \)
Let \( z = \sin^{-1} \frac{1}{\sqrt{5}} \). Then:
\[ \sin z = \frac{1}{\sqrt{5}} \]
Using the Pythagorean identity, we can find \( \cos z \):
\[ \cos z = \sqrt{1 - \sin^2 z} = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

### Step 4: Find \( \tan z \)
Now, we can find \( \tan z \):
\[ \tan z = \frac{\sin z}{\cos z} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]
Thus, we have:
\[ z = \tan^{-1} \frac{1}{2} \]

### Step 5: Substitute back into the equation
Now substituting back, we have:
\[ y = \frac{\pi}{4} - \tan^{-1} \frac{1}{2} \]

### Step 6: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan\left(\frac{\pi}{4} - \tan^{-1} \frac{1}{2}\right) = \frac{1 - \tan(\tan^{-1} \frac{1}{2})}{1 + \tan(\tan^{-1} \frac{1}{2})} = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \]
Thus, we have:
\[ \tan y = \frac{1}{3} \]

### Step 7: Relate back to \( x \)
Since \( y = \cot^{-1} x \), we have:
\[ \tan y = \frac{1}{x} \implies \frac{1}{3} = \frac{1}{x} \implies x = 3 \]

### Conclusion
The solution to the equation \( \cot^{-1} x + \sin^{-1} \frac{1}{\sqrt{5}} = \frac{\pi}{4} \) is:
\[ \boxed{3} \]