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 Fraction
Multiply both sides of the equation by 2 to eliminate the fraction:
\[ \sin^{-1}\left(\frac{3 \sin 2\theta}{5 + 4 \cos 2\theta}\right) = \frac{\pi}{2} \]

### Step 2: Apply the Sine Function
Taking the sine of both sides gives:
\[ \frac{3 \sin 2\theta}{5 + 4 \cos 2\theta} = \sin\left(\frac{\pi}{2}\right) \]
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:
\[ \frac{3 \sin 2\theta}{5 + 4 \cos 2\theta} = 1 \]

### Step 3: Cross Multiply
Cross multiplying gives:
\[ 3 \sin 2\theta = 5 + 4 \cos 2\theta \]

### Step 4: Substitute for \(\sin 2\theta\) and \(\cos 2\theta\)
Using the double angle identities:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \quad \text{and} \quad \cos 2\theta = 1 - 2 \sin^2 \theta \]
Substituting these into the equation:
\[ 3(2 \sin \theta \cos \theta) = 5 + 4(1 - 2 \sin^2 \theta) \]
This simplifies to:
\[ 6 \sin \theta \cos \theta = 5 + 4 - 8 \sin^2 \theta \]
\[ 6 \sin \theta \cos \theta = 9 - 8 \sin^2 \theta \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 8 \sin^2 \theta + 6 \sin \theta \cos \theta - 9 = 0 \]

### Step 6: Substitute \(\tan \theta\)
Let \( x = \tan \theta \), then \( \sin \theta = \frac{x}{\sqrt{1+x^2}} \) and \( \cos \theta = \frac{1}{\sqrt{1+x^2}} \):
\[ 6 \cdot \frac{2x}{1+x^2} = 9 - 8 \cdot \frac{x^2}{1+x^2} \]
This leads to:
\[ \frac{12x}{1+x^2} = 9 - \frac{8x^2}{1+x^2} \]
Multiplying through by \( 1+x^2 \):
\[ 12x = 9(1+x^2) - 8x^2 \]
\[ 12x = 9 + 9x^2 - 8x^2 \]
\[ 12x = 9 + x^2 \]

### Step 7: Rearranging to Form a Quadratic Equation
Rearranging gives:
\[ x^2 - 12x + 9 = 0 \]

### Step 8: Solving the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]
\[ x = \frac{12 \pm \sqrt{144 - 36}}{2} \]
\[ x = \frac{12 \pm \sqrt{108}}{2} \]
\[ x = \frac{12 \pm 6\sqrt{3}}{2} \]
\[ x = 6 \pm 3\sqrt{3} \]

### Step 9: Conclusion
Thus, \( \tan \theta = 6 + 3\sqrt{3} \) or \( \tan \theta = 6 - 3\sqrt{3} \).

### Final Answer
The value of \( \tan \theta \) is \( 6 + 3\sqrt{3} \) or \( 6 - 3\sqrt{3} \).

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