If `((x+1)^(2))/(x(x^(2)+1))=(A)/(x)+(Bx+C)/(x^(2)+1)` then `cos^(-1)(A/C)=`

Options

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

Answer

To solve the problem step by step, we start with the equation given in the question:

\[ \frac{(x+1)^2}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \]

### Step 1: Rewrite the Left Side
First, we simplify the left-hand side:

\[ \frac{(x+1)^2}{x(x^2+1)} = \frac{x^2 + 2x + 1}{x(x^2 + 1)} \]

### Step 2: Combine the Right Side
Next, we need to combine the right-hand side into a single fraction:

\[ \frac{A}{x} + \frac{Bx+C}{x^2+1} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)} \]

### Step 3: Expand the Right Side
Now, we expand the numerator on the right side:

\[ A(x^2 + 1) + (Bx + C)x = Ax^2 + A + Bx^2 + Cx = (A + B)x^2 + Cx + A \]

### Step 4: Set the Numerators Equal
Now we set the numerators of both sides equal to each other:

\[ x^2 + 2x + 1 = (A + B)x^2 + Cx + A \]

### Step 5: Compare Coefficients
We will compare coefficients for \(x^2\), \(x\), and the constant term:

1. Coefficient of \(x^2\): \(1 = A + B\)
2. Coefficient of \(x\): \(2 = C\)
3. Constant term: \(1 = A\)

### Step 6: Solve for A, B, and C
From the third equation, we have:

\[ A = 1 \]

Substituting \(A = 1\) into the first equation:

\[ 1 = 1 + B \implies B = 0 \]

From the second equation, we have:

\[ C = 2 \]

### Step 7: Find \( \cos^{-1} \left( \frac{A}{C} \right) \)
Now we need to find \( \cos^{-1} \left( \frac{A}{C} \right) \):

\[ \frac{A}{C} = \frac{1}{2} \]

### Step 8: Determine the Angle
The angle whose cosine is \( \frac{1}{2} \) is:

\[ \theta = \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \]

### Final Answer
Thus, the final answer is:

\[ \cos^{-1} \left( \frac{A}{C} \right) = \frac{\pi}{3} \]

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