If `cot^(-1) x+ sin^(-1)""1/sqrt5 = pi/4 , " then " x=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \cot^{-1} x = \tan^{-1} \left( \frac{1}{x} \right) \). Therefore, we can rewrite the equation as:
\[ \tan^{-1} \left( \frac{1}{x} \right) + \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{\pi}{4} \]

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

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

### Step 4: Calculate \( \tan \theta \)
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]

### Step 5: Substitute back into the equation
Now we substitute \( \tan \theta \) back into our equation:
\[ \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{4} - \tan^{-1} \left( \frac{1}{2} \right) \]

### Step 6: Use the tangent subtraction formula
Using the tangent subtraction formula:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
where \( a = \frac{\pi}{4} \) and \( b = \tan^{-1} \left( \frac{1}{2} \right) \):
\[ \tan \left( \frac{\pi}{4} - \tan^{-1} \left( \frac{1}{2} \right) \right) = \frac{1 - \frac{1}{2}}{1 + 1 \cdot \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \]

### Step 7: Set the equation for \( \frac{1}{x} \)
Thus, we have:
\[ \frac{1}{x} = \frac{1}{3} \]

### Step 8: Solve for \( x \)
Taking the reciprocal gives:
\[ x = 3 \]

### Final Answer
The value of \( x \) is:
\[ \boxed{3} \]