If `1/2 sin^(-1)((3sin2theta)/(5+4cos2theta))=(pi)/4`, then `tan theta` is equal to

Explanation

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

### Step 1: Eliminate the inverse sine function
Multiply both sides by 2:
\[ \sin^{-1}\left(\frac{3 \sin 2\theta}{5 + 4 \cos 2\theta}\right) = \frac{\pi}{2} \]

### Step 2: Apply the sine function
Taking the sine of both sides gives:
\[ \frac{3 \sin 2\theta}{5 + 4 \cos 2\theta} = 1 \]

### Step 3: Rearrange the equation
Multiply both sides by \(5 + 4 \cos 2\theta\):
\[ 3 \sin 2\theta = 5 + 4 \cos 2\theta \]

### Step 4: Use trigonometric identities
Recall the identities:
- \( \sin 2\theta = 2 \sin \theta \cos \theta \)
- \( \cos 2\theta = 1 - 2 \sin^2 \theta \)

Substituting these identities into the equation gives:
\[ 3(2 \sin \theta \cos \theta) = 5 + 4(1 - 2 \sin^2 \theta) \]

### Step 5: Simplify the equation
Expanding the right side:
\[ 6 \sin \theta \cos \theta = 5 + 4 - 8 \sin^2 \theta \]
\[ 6 \sin \theta \cos \theta = 9 - 8 \sin^2 \theta \]

### Step 6: Rearranging terms
Rearranging gives:
\[ 8 \sin^2 \theta + 6 \sin \theta \cos \theta - 9 = 0 \]

### Step 7: Substitute \( \tan \theta \)
Let \( \tan \theta = y \), then \( \sin \theta = \frac{y}{\sqrt{1+y^2}} \) and \( \cos \theta = \frac{1}{\sqrt{1+y^2}} \). Substitute these into the equation:
\[ 8 \left(\frac{y^2}{1+y^2}\right) + 6 \left(\frac{y}{\sqrt{1+y^2}} \cdot \frac{1}{\sqrt{1+y^2}}\right) - 9 = 0 \]
This simplifies to:
\[ 8y^2 + 6y - 9(1+y^2) = 0 \]
\[ 8y^2 + 6y - 9 - 9y^2 = 0 \]
\[ -y^2 + 6y - 9 = 0 \]

### Step 8: Solve the quadratic equation
Multiplying through by -1 gives:
\[ y^2 - 6y + 9 = 0 \]
Factoring:
\[ (y - 3)^2 = 0 \]
Thus, \( y = 3 \).

### Step 9: Conclusion
Since \( y = \tan \theta \), we have:
\[ \tan \theta = 3 \]

### Final Answer
Thus, the value of \( \tan \theta \) is \( 3 \).

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