If `sin^(-1) ((3 sin 2 theta)/(5+4 cos 2 theta))=2 tan^(-1)x`, then `x=`

Options

A. `tan 3 theta`
B. `3 tan theta`
C. `1/3 tan theta`
D. `3 cot theta`

Answer

To solve the equation \( \sin^{-1} \left( \frac{3 \sin 2\theta}{5 + 4 \cos 2\theta} \right) = 2 \tan^{-1} x \), we can follow these steps:

### Step 1: Rewrite the left-hand side
We start with the expression inside the inverse sine function:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
and
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = 1 - 2\sin^2 \theta = 2\cos^2 \theta - 1 \]
Thus, we can express \( \sin 2\theta \) and \( \cos 2\theta \) in terms of \( \tan \theta \).

### Step 2: Substitute \( \sin 2\theta \) and \( \cos 2\theta \)
Using the double angle formulas:
\[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \]
and
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
we can rewrite the expression:
\[ \frac{3 \sin 2\theta}{5 + 4 \cos 2\theta} = \frac{3 \cdot \frac{2\tan \theta}{1 + \tan^2 \theta}}{5 + 4 \cdot \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} \]

### Step 3: Simplify the denominator
The denominator becomes:
\[ 5 + 4 \cdot \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{5(1 + \tan^2 \theta) + 4(1 - \tan^2 \theta)}{1 + \tan^2 \theta} = \frac{5 + 5\tan^2 \theta + 4 - 4\tan^2 \theta}{1 + \tan^2 \theta} = \frac{9 + \tan^2 \theta}{1 + \tan^2 \theta} \]

### Step 4: Substitute back into the expression
Now we have:
\[ \frac{3 \cdot \frac{2\tan \theta}{1 + \tan^2 \theta}}{\frac{9 + \tan^2 \theta}{1 + \tan^2 \theta}} = \frac{6\tan \theta}{9 + \tan^2 \theta} \]

### Step 5: Set the equation
Now we set the equation:
\[ \sin^{-1} \left( \frac{6\tan \theta}{9 + \tan^2 \theta} \right) = 2 \tan^{-1} x \]

### Step 6: Use the identity for \( \tan^{-1} \)
Using the identity \( \tan(2A) = \frac{2\tan A}{1 - \tan^2 A} \), we can express \( \sin^{-1} y = 2 \tan^{-1} x \) as:
\[ \sin(2 \tan^{-1} x) = \frac{2x}{1+x^2} \]
Thus, we equate:
\[ \frac{6\tan \theta}{9 + \tan^2 \theta} = \frac{2x}{1+x^2} \]

### Step 7: Cross-multiply
Cross-multiplying gives:
\[ 6\tan \theta(1 + x^2) = 2x(9 + \tan^2 \theta) \]

### Step 8: Solve for \( x \)
Rearranging and solving for \( x \) gives:
\[ x = \frac{3\tan \theta}{9 - 3\tan \theta} \]

### Conclusion
Thus, the value of \( x \) is:
\[ x = \frac{3\tan \theta}{9 - 3\tan \theta} \]