A solution of the equation `cot^(-1)2=cot^(-1)x+cot^(-1)(10-x)" where " 1 lt x lt 9` is : (a) 7 (b) 3 (c) 2 (d) 5

Options

A. 7
B. 3
C. 2
D. 5

Answer

To solve the equation \( \cot^{-1}(2) = \cot^{-1}(x) + \cot^{-1}(10 - x) \) where \( 1 < x < 9 \), we can follow these steps:

### Step 1: Use the identity for the sum of inverse cotangents
We know the identity:
\[ \cot^{-1}(A) + \cot^{-1}(B) = \cot^{-1}\left(\frac{AB - 1}{A + B}\right) \]
Let \( A = x \) and \( B = 10 - x \). Then, we can rewrite the equation as:
\[ \cot^{-1}(2) = \cot^{-1}\left(\frac{x(10 - x) - 1}{x + (10 - x)}\right) \]

### Step 2: Simplify the right-hand side
The denominator simplifies to:
\[ x + (10 - x) = 10 \]
So, we have:
\[ \cot^{-1}(2) = \cot^{-1}\left(\frac{x(10 - x) - 1}{10}\right) \]

### Step 3: Apply the cotangent function
Taking the cotangent of both sides gives:
\[ 2 = \frac{x(10 - x) - 1}{10} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying yields:
\[ 20 = x(10 - x) - 1 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ x(10 - x) = 21 \]
This simplifies to:
\[ 10x - x^2 = 21 \]
Rearranging further, we get:
\[ x^2 - 10x + 21 = 0 \]

### Step 6: Solve the quadratic equation
We can factor the quadratic equation:
\[ (x - 7)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 7 \quad \text{or} \quad x = 3 \]

### Step 7: Check the range of x
Since \( 1 < x < 9 \), both \( x = 7 \) and \( x = 3 \) are valid solutions.

### Conclusion
The solutions to the equation are \( x = 7 \) and \( x = 3 \). Therefore, both options (a) 7 and (b) 3 are correct.

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