Prove that `tan^(-1)(x+1)+tan^(-1)(x-1)=tan^(-1)((2x)/(2-x^2))`

Options

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

Answer

To prove that

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

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

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

provided that \(ab < 1\).

### Step 1: Identify \(a\) and \(b\)

Let:
- \(a = x + 1\)
- \(b = x - 1\)

### Step 2: Calculate \(a + b\)

\[ a + b = (x + 1) + (x - 1) = 2x \]

### Step 3: Calculate \(ab\)

\[ ab = (x + 1)(x - 1) = x^2 - 1 \]

### Step 4: Check the condition for the identity

We need to ensure that \(ab < 1\):

\[ x^2 - 1 < 1 \implies x^2 < 2 \implies |x| < \sqrt{2} \]

This condition will be satisfied for \(x\) values within the interval \((- \sqrt{2}, \sqrt{2})\).

### Step 5: Apply the identity

Using the identity:

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

### Step 6: Simplify the denominator

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

### Step 7: Substitute back into the equation

Thus, we have:

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

### Conclusion

This shows that:

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

Hence, we have proved the required equation.

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