The value (s) of `theta` satisfying the equation `theta=tan^(- 1)(2tan^2theta)-1/2(sin^(- 1)((3sin2theta)/(5+4cos2theta)))` is

Options

A. (a) `npi+pi/4`
B. (b) `npi+tan^(-1)(-2)`
C. (c) `npi+pi/3`
D. (d) `npi+pi/6`

Answer

To solve the equation

\[ \theta = \tan^{-1}(2\tan^2\theta) - \frac{1}{2}\sin^{-1}\left(\frac{3\sin 2\theta}{5 + 4\cos 2\theta}\right) \]

we will follow these steps:

### Step 1: Multiply both sides by 2
Multiply the entire equation by 2 to simplify the expression:

\[ 2\theta = 2\tan^{-1}(2\tan^2\theta) - \sin^{-1}\left(\frac{3\sin 2\theta}{5 + 4\cos 2\theta}\right) \]

### Step 2: Substitute \(\theta = \frac{\pi}{4}\)
Let's check if \(\theta = \frac{\pi}{4}\) satisfies the equation.

Calculating the left-hand side (LHS):

\[ 2\theta = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

Now, calculate the right-hand side (RHS):

\[ RHS = 2\tan^{-1}(2\tan^2(\frac{\pi}{4})) - \sin^{-1}\left(\frac{3\sin(2 \cdot \frac{\pi}{4})}{5 + 4\cos(2 \cdot \frac{\pi}{4})}\right) \]

### Step 3: Calculate \(\tan^2(\frac{\pi}{4})\)
Since \(\tan(\frac{\pi}{4}) = 1\):

\[ \tan^2(\frac{\pi}{4}) = 1 \]

Thus,

\[ 2\tan^2(\frac{\pi}{4}) = 2 \cdot 1 = 2 \]

Now, calculate \(2\tan^{-1}(2)\):

\[ RHS = 2\tan^{-1}(2) - \sin^{-1}\left(\frac{3\sin(\frac{\pi}{2})}{5 + 4\cos(\frac{\pi}{2})}\right) \]

### Step 4: Calculate \(\sin(\frac{\pi}{2})\) and \(\cos(\frac{\pi}{2})\)
We know:

\[ \sin(\frac{\pi}{2}) = 1 \quad \text{and} \quad \cos(\frac{\pi}{2}) = 0 \]

Thus,

\[ RHS = 2\tan^{-1}(2) - \sin^{-1}\left(\frac{3 \cdot 1}{5 + 0}\right) = 2\tan^{-1}(2) - \sin^{-1}\left(\frac{3}{5}\right) \]

### Step 5: Use the identity for \(2\tan^{-1}(x)\)
Using the identity \(2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\):

Let \(x = 2\):

\[ 2\tan^{-1}(2) = \tan^{-1}\left(\frac{4}{1-4}\right) = \tan^{-1}\left(-\frac{4}{3}\right) \]

### Step 6: Combine the terms
Now we have:

\[ RHS = \tan^{-1}\left(-\frac{4}{3}\right) - \sin^{-1}\left(\frac{3}{5}\right) \]

### Step 7: Use the addition formula for inverse tangents
Using the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]

We can express:

\[ RHS = -\tan^{-1}\left(\frac{4}{3}\right) - \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 8: Check if LHS equals RHS
Now we check if:

\[ \frac{\pi}{2} = \tan^{-1}\left(\text{undefined}\right) \]

This indicates that the equation holds true when \(\theta = \frac{\pi}{4}\).

### Conclusion
Thus, the values of \(\theta\) satisfying the equation are:

\[ \theta = n\pi + \frac{\pi}{4} \]

where \(n\) is any integer.