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

Explanation

To solve the problem, we need to analyze the expression \( z = \frac{3 + i \sin \theta}{4 - i \cos \theta} \) and determine the conditions under which it is purely real. We will then find the argument of \( \sin \theta + i \cos \theta \).

### Step 1: Rationalize the denominator
We start by multiplying the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{(3 + i \sin \theta)(4 + i \cos \theta)}{(4 - i \cos \theta)(4 + i \cos \theta)} \]

### Step 2: Calculate the denominator
The denominator simplifies as follows:

\[ (4 - i \cos \theta)(4 + i \cos \theta) = 4^2 - (i \cos \theta)^2 = 16 + \cos^2 \theta \]

### Step 3: Calculate the numerator
Now, we calculate the numerator:

\[ (3 + i \sin \theta)(4 + i \cos \theta) = 3 \cdot 4 + 3 \cdot i \cos \theta + i \sin \theta \cdot 4 + i^2 \sin \theta \cos \theta \]
\[ = 12 + 3i \cos \theta + 4i \sin \theta - \sin \theta \cos \theta \]
\[ = (12 - \sin \theta \cos \theta) + i(3 \cos \theta + 4 \sin \theta) \]

### Step 4: Combine the results
Now we can write \( z \) as:

\[ z = \frac{(12 - \sin \theta \cos \theta) + i(3 \cos \theta + 4 \sin \theta)}{16 + \cos^2 \theta} \]

### Step 5: Determine when \( z \) is purely real
For \( z \) to be purely real, the imaginary part must be zero:

\[ 3 \cos \theta + 4 \sin \theta = 0 \]

### Step 6: Solve for \( \tan \theta \)
From the equation \( 3 \cos \theta + 4 \sin \theta = 0 \), we can rearrange it to find:

\[ 4 \sin \theta = -3 \cos \theta \implies \tan \theta = -\frac{3}{4} \]

### Step 7: Find the argument of \( \sin \theta + i \cos \theta \)
We need to find the argument of \( \sin \theta + i \cos \theta \). The argument is given by:

\[ \arg(\sin \theta + i \cos \theta) = \tan^{-1}\left(\frac{\cos \theta}{\sin \theta}\right) \]

### Step 8: Substitute \( \tan \theta \)
Since \( \tan \theta = -\frac{3}{4} \), we can find \( \cot \theta \):

\[ \cot \theta = -\frac{4}{3} \]

### Step 9: Determine the quadrant
Given that \( \theta \in \left(\frac{\pi}{2}, \pi\right) \), \( \sin \theta > 0 \) and \( \cos \theta < 0 \). Therefore, \( \sin \theta + i \cos \theta \) lies in the second quadrant.

### Step 10: Final argument calculation
The argument in the second quadrant is given by:

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

Thus, the final answer is:

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