The values of x satisfying
`"tan"^(-1) (x+3) -"tan"^(-1) (x-3) = "sin"^(-1)((3)/(5))` are

Options

A. `+-4`
B. 0, 4
C. `-4, 0`
D. 4, 5

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: Rewrite the equation using the tangent subtraction formula
We can use the formula for the difference of two inverse tangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
Let \( a = x + 3 \) and \( b = x - 3 \). Thus, we have:
\[ \tan^{-1}(x+3) - \tan^{-1}(x-3) = \tan^{-1}\left(\frac{(x+3) - (x-3)}{1 + (x+3)(x-3)}\right) \]
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 the right-hand side
Now we need to equate this to \( \sin^{-1}\left(\frac{3}{5}\right) \). First, we need to find the value of \( \tan\left(\sin^{-1}\left(\frac{3}{5}\right)\right) \):
\[ \sin(\theta) = \frac{3}{5} \implies \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus,
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{4/5} = \frac{3}{4} \]
Therefore, we have:
\[ \tan^{-1}\left(\frac{6}{x^2 - 8}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

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

### 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: Find the values of \( x \)
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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