Number of solutions of the equation `tan^(-1)((1)/(a-1))=tan^(-1)((1)/(x))+tan^(-1)((1)/(a^(2)-x+1))` is :

Options

A. One
B. Two
C. Three
D. zero

Answer

To solve the equation

\[ \tan^{-1}\left(\frac{1}{a-1}\right) = \tan^{-1}\left(\frac{1}{x}\right) + \tan^{-1}\left(\frac{1}{a^2 - x + 1}\right), \]

we can use the identity for the sum of inverse tangents:

\[ \tan^{-1}(u) + \tan^{-1}(v) = \tan^{-1}\left(\frac{u + v}{1 - uv}\right), \]

provided that \(uv < 1\).

### Step 1: Apply the identity

Let \(u = \frac{1}{x}\) and \(v = \frac{1}{a^2 - x + 1}\). Then, we can rewrite the right-hand side:

\[ \tan^{-1}\left(\frac{1}{x}\right) + \tan^{-1}\left(\frac{1}{a^2 - x + 1}\right) = \tan^{-1}\left(\frac{\frac{1}{x} + \frac{1}{a^2 - x + 1}}{1 - \frac{1}{x(a^2 - x + 1)}}\right). \]

### Step 2: Simplify the right-hand side

Calculating \(u + v\):

\[ u + v = \frac{1}{x} + \frac{1}{a^2 - x + 1} = \frac{(a^2 - x + 1) + x}{x(a^2 - x + 1)} = \frac{a^2 + 1}{x(a^2 - x + 1)}. \]

Now, calculating \(1 - uv\):

\[ uv = \frac{1}{x(a^2 - x + 1)} \quad \text{thus} \quad 1 - uv = 1 - \frac{1}{x(a^2 - x + 1)} = \frac{x(a^2 - x + 1) - 1}{x(a^2 - x + 1)}. \]

### Step 3: Combine the results

Now we can substitute these into the identity:

\[ \tan^{-1}\left(\frac{\frac{a^2 + 1}{x(a^2 - x + 1)}}{\frac{x(a^2 - x + 1) - 1}{x(a^2 - x + 1)}}\right) = \tan^{-1}\left(\frac{(a^2 + 1)}{x(a^2 - x + 1) - 1}\right). \]

### Step 4: Set the two sides equal

Now we have:

\[ \tan^{-1}\left(\frac{1}{a-1}\right) = \tan^{-1}\left(\frac{a^2 + 1}{x(a^2 - x + 1) - 1}\right). \]

Taking the tangent of both sides gives:

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

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

\[ 1 \cdot (x(a^2 - x + 1) - 1) = (a^2 + 1)(a - 1). \]

### Step 6: Solve for \(x\)

This leads to a quadratic equation in \(x\). The number of solutions to this equation can be determined by the discriminant.

### Step 7: Determine the number of solutions

The discriminant of the quadratic equation will tell us how many real solutions exist. If the discriminant is positive, there are two solutions; if it is zero, there is one solution; and if it is negative, there are no solutions.

### Conclusion

Thus, the number of solutions to the original equation is determined by the discriminant of the resulting quadratic equation.

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