If `"sin"^(-1)((6x)/(1+9x^2))=2 "tan"^(-1)(ax)`, then a=

Options

A. 3
B. 8
C. 6
D. 9

Answer

To solve the equation \( \sin^{-1}\left(\frac{6x}{1+9x^2}\right) = 2 \tan^{-1}(ax) \), we will follow these steps:

### Step 1: Simplify the Left-Hand Side
We can rewrite the left-hand side using the double angle formula for sine. We know that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
We can express \( \frac{6x}{1 + 9x^2} \) in terms of \( \tan \) to use this identity.

### Step 2: Set \( 3x = \tan(\theta) \)
Let \( 3x = \tan(\theta) \). Then, we can express \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \sin(\theta) = \frac{3x}{\sqrt{1 + (3x)^2}} = \frac{3x}{\sqrt{1 + 9x^2}} \]
\[ \cos(\theta) = \frac{1}{\sqrt{1 + (3x)^2}} = \frac{1}{\sqrt{1 + 9x^2}} \]

### Step 3: Substitute into the Sine Double Angle Formula
Using the double angle formula:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{3x}{\sqrt{1 + 9x^2}}\right) \left(\frac{1}{\sqrt{1 + 9x^2}}\right) = \frac{6x}{1 + 9x^2} \]
Thus, we have:
\[ \sin^{-1}\left(\frac{6x}{1 + 9x^2}\right) = 2\theta \]

### Step 4: Relate \( \theta \) to \( \tan^{-1} \)
Since we set \( \tan(\theta) = 3x \), we have:
\[ \theta = \tan^{-1}(3x) \]
Thus:
\[ 2\theta = 2\tan^{-1}(3x) \]

### Step 5: Set the Two Sides Equal
Now we can equate both sides:
\[ 2\tan^{-1}(3x) = 2\tan^{-1}(ax) \]

### Step 6: Divide by 2
Dividing both sides by 2 gives:
\[ \tan^{-1}(3x) = \tan^{-1}(ax) \]

### Step 7: Equate the Arguments
Since the inverse tangent function is one-to-one, we can equate the arguments:
\[ 3x = ax \]

### Step 8: Solve for \( a \)
Assuming \( x \neq 0 \), we can divide both sides by \( x \):
\[ 3 = a \]

### Conclusion
Thus, the value of \( a \) is:
\[ \boxed{3} \]