If `tan^(-1) (x+3)- tan^(-1) (x-3) = sin^(-1) ((3)/(5))` then the value of x is

Options

A. 3
B. 14
C. ` pm4`
D. 7

Answer

To solve the equation \( \tan^{-1}(x+3) - \tan^{-1}(x-3) = \sin^{-1}\left(\frac{3}{5}\right) \), we can follow these steps:

### Step 1: Use the formula for the difference of inverse tangents
We know that:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
Applying this to our equation:
\[ \tan^{-1}(x+3) - \tan^{-1}(x-3) = \tan^{-1}\left(\frac{(x+3) - (x-3)}{1 + (x+3)(x-3)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{6}{1 + (x^2 - 9)}\right) = \tan^{-1}\left(\frac{6}{x^2 - 8}\right) \]

### Step 2: Set the equation equal to \(\sin^{-1}\left(\frac{3}{5}\right)\)
We know that:
\[ \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]
Thus, we can rewrite our equation as:
\[ \tan^{-1}\left(\frac{6}{x^2 - 8}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 3: Equate the arguments of the tangent inverse
Since the tangent function is one-to-one, we can set the arguments equal:
\[ \frac{6}{x^2 - 8} = \frac{3}{4} \]

### Step 4: Cross-multiply to solve for \(x\)
Cross-multiplying gives us:
\[ 6 \cdot 4 = 3 \cdot (x^2 - 8) \]
This simplifies to:
\[ 24 = 3x^2 - 24 \]

### Step 5: Rearrange the equation
Adding 24 to both sides:
\[ 24 + 24 = 3x^2 \]
\[ 48 = 3x^2 \]

### Step 6: Solve for \(x^2\)
Dividing both sides by 3:
\[ x^2 = \frac{48}{3} = 16 \]

### Step 7: Take the square root
Taking the square root of both sides gives:
\[ x = \pm 4 \]

### Conclusion
Thus, the values of \(x\) are:
\[ x = 4 \quad \text{or} \quad x = -4 \]