If z = `(3+isintheta)/(4-icostheta)` is purely real and `theta in (pi/2,pi)` ,then `arg(sintheta + i costheta)` is -

Options

A. `pi-tan^(-1).((3)/(4))`
B. `pi-tan^(-1)((4)/(3))`
C. `tan^(-1)((4)/(3))`
D. `-tan^(-1)((3)/(4))`

Answer

To solve the problem, we need to find the argument of \( \sin \theta + i \cos \theta \) given that \( z = \frac{3 + i \sin \theta}{4 - i \cos \theta} \) is purely real and \( \theta \in \left( \frac{\pi}{2}, \pi \right) \).

### Step-by-Step Solution:

1. **Multiply by the Conjugate**:
We start with the expression for \( z \):
\[ z = \frac{3 + i \sin \theta}{4 - i \cos \theta} \]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(3 + i \sin \theta)(4 + i \cos \theta)}{(4 - i \cos \theta)(4 + i \cos \theta)} \]

2. **Simplify the Denominator**:
The denominator simplifies to:
\[ (4 - i \cos \theta)(4 + i \cos \theta) = 4^2 + \cos^2 \theta = 16 + \cos^2 \theta \]

3. **Expand the Numerator**:
The numerator expands to:
\[ (3 + i \sin \theta)(4 + i \cos \theta) = 12 + 3i \cos \theta + 4i \sin \theta - \sin \theta \cos \theta \]
Thus, we have:
\[ z = \frac{12 - \sin \theta \cos \theta + i(3 \cos \theta + 4 \sin \theta)}{16 + \cos^2 \theta} \]

4. **Separate Real and Imaginary Parts**:
The real part \( R \) and imaginary part \( I \) of \( z \) are:
\[ R = \frac{12 - \sin \theta \cos \theta}{16 + \cos^2 \theta}, \quad I = \frac{3 \cos \theta + 4 \sin \theta}{16 + \cos^2 \theta} \]

5. **Set Imaginary Part to Zero**:
Since \( z \) is purely real, we set the imaginary part \( I \) to zero:
\[ 3 \cos \theta + 4 \sin \theta = 0 \]
This implies:
\[ 3 \cos \theta = -4 \sin \theta \]
Dividing both sides by \( \cos \theta \) (noting \( \cos \theta \neq 0 \) in the given interval):
\[ 3 = -4 \tan \theta \quad \Rightarrow \quad \tan \theta = -\frac{3}{4} \]

6. **Find the Argument**:
We need to find the argument of \( \sin \theta + i \cos \theta \). We can express this as:
\[ \arg(\sin \theta + i \cos \theta) = \tan^{-1}\left(\frac{\cos \theta}{\sin \theta}\right) \]
Since \( \tan \theta = -\frac{3}{4} \), we have:
\[ \frac{\cos \theta}{\sin \theta} = \frac{1}{\tan \theta} = -\frac{4}{3} \]
Therefore:
\[ \arg(\sin \theta + i \cos \theta) = \tan^{-1}\left(-\frac{4}{3}\right) \]

7. **Determine the Quadrant**:
Since \( \theta \in \left( \frac{\pi}{2}, \pi \right) \), both \( \sin \theta \) and \( \cos \theta \) are positive and negative, respectively. Thus, the argument is in the second quadrant:
\[ \arg(\sin \theta + i \cos \theta) = \pi - \tan^{-1}\left(\frac{4}{3}\right) \]

### Final Answer:
Thus, the argument of \( \sin \theta + i \cos \theta \) is:
\[ \arg(\sin \theta + i \cos \theta) = \pi - \tan^{-1}\left(\frac{4}{3}\right) \]