The value(s) of `x` satisfying `tan^(-1)(x+3)-tan^(-1)(x-3)=sin^(-1)(3/5)` may be

Options

A. `-2`
B. `-1`
C. 0
D. 2

Explanation

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

### Step 1: Evaluate \( \sin^{-1}\left(\frac{3}{5}\right) \)
We know that \( \sin^{-1}\left(\frac{3}{5}\right) \) gives us an angle \( \theta \) such that \( \sin(\theta) = \frac{3}{5} \). We can find the corresponding value of \( \tan(\theta) \) using the Pythagorean theorem.

Let \( \sin(\theta) = \frac{3}{5} \). Then, we can find \( \cos(\theta) \):
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Now, we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 2: Use the tangent subtraction formula
The formula for the difference of two inverse tangents is:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
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:
\[ \tan^{-1}\left(\frac{6}{1 + (x^2 - 9)}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 3: Set the arguments equal
Since the tangent function is one-to-one, we can set the arguments equal:
\[ \frac{6}{1 + x^2 - 9} = \frac{3}{4} \]
This simplifies to:
\[ \frac{6}{x^2 - 8} = \frac{3}{4} \]

### Step 4: Cross-multiply and solve for \( x^2 \)
Cross-multiplying gives:
\[ 6 \cdot 4 = 3(x^2 - 8) \]
\[ 24 = 3x^2 - 24 \]
Adding 24 to both sides:
\[ 48 = 3x^2 \]
Dividing by 3:
\[ x^2 = 16 \]

### Step 5: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm 4 \]

### Final Answer
The values of \( x \) satisfying the equation are:
\[ x = 4 \quad \text{and} \quad x = -4 \]

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