If vector ` vec b=(t a nalpha,-1,2sqrt(sinalpha//2))a n d vec c=(t a nalpha, t a nalpha, -3/(sqrt(sinalpha//2)))` are orthogonal and vector ` vec a=(1,3,sin2alpha)` makes an obtuse angle with the z-axis, then the value of `alpha` is `a. alpha=(4n+1)pi+tan^(-1)2` `b. alpha=(4n+1)pi-tan^(-1)2` `c. alpha=(4n+2)pi+tan^(-1)2` `d. alpha=(4n+2)pi-tan^(-1)2`

Options

A. `alpha= ( 4n+1 ) pi + tan^(-1) 2`
B. `alpha= ( 4n+1 ) pi - tan^(-1) 2`
C. `alpha= ( 4n+2 ) pi + tan^(-1) 2`
D. `alpha= ( 4n+2 ) pi - tan^(-1) 2`

Explanation

To solve the problem step by step, we will analyze the given vectors and their properties.

### Step 1: Define the vectors
We have:
- \( \vec{b} = (\tan \alpha, -1, 2\sqrt{\frac{\sin \alpha}{2}}) \)
- \( \vec{c} = (\tan \alpha, \tan \alpha, -\frac{3}{\sqrt{\frac{\sin \alpha}{2}}}) \)
- \( \vec{a} = (1, 3, \sin 2\alpha) \)

### Step 2: Use the orthogonality condition
Since vectors \( \vec{b} \) and \( \vec{c} \) are orthogonal, their dot product must equal zero:
\[ \vec{b} \cdot \vec{c} = 0 \]
Calculating the dot product:
\[ (\tan \alpha)(\tan \alpha) + (-1)(\tan \alpha) + \left(2\sqrt{\frac{\sin \alpha}{2}}\right)\left(-\frac{3}{\sqrt{\frac{\sin \alpha}{2}}}\right) = 0 \]
This simplifies to:
\[ \tan^2 \alpha - \tan \alpha - 6 = 0 \]

### Step 3: Solve the quadratic equation
Now we can factor the quadratic:
\[ (\tan \alpha - 3)(\tan \alpha + 2) = 0 \]
Thus, we have two solutions:
\[ \tan \alpha = 3 \quad \text{or} \quad \tan \alpha = -2 \]

### Step 4: Find the angles
From \( \tan \alpha = 3 \):
\[ \alpha = n\pi + \tan^{-1}(3) \]
From \( \tan \alpha = -2 \):
\[ \alpha = n\pi - \tan^{-1}(2) \]

### Step 5: Analyze the angle with the z-axis
The vector \( \vec{a} \) makes an obtuse angle with the z-axis, which means:
\[ \vec{a} \cdot \hat{k} < 0 \]
Calculating the dot product with the unit vector in the z-direction \( \hat{k} = (0, 0, 1) \):
\[ \sin 2\alpha < 0 \]
This implies:
\[ 2\alpha \in (2m-1)\pi \text{ to } 2m\pi \quad \Rightarrow \quad \alpha \in \left(\frac{(2m-1)\pi}{2}, m\pi\right) \]
This indicates that \( \alpha \) is in the second or fourth quadrant.

### Step 6: Find the intersection of ranges
We need to find the intersection of the ranges from the solutions of \( \tan \alpha \) and the angle condition:
- From \( \tan \alpha = 3 \): \( \alpha = n\pi + \tan^{-1}(3) \)
- From \( \tan \alpha = -2 \): \( \alpha = n\pi - \tan^{-1}(2) \)

### Step 7: Determine the valid solutions
Considering the ranges:
1. \( n\pi + \tan^{-1}(3) \) does not satisfy the obtuse angle condition.
2. \( n\pi - \tan^{-1}(2) \) can satisfy the condition for certain values of \( n \).

Thus, we conclude:
\[ \alpha = n\pi - \tan^{-1}(2) \]
This can be expressed in the forms:
- \( \alpha = (4n + 1)\pi - \tan^{-1}(2) \)
- \( \alpha = (4n + 2)\pi - \tan^{-1}(2) \)

### Final Answer
The values of \( \alpha \) are:
- \( \alpha = (4n + 1)\pi - \tan^{-1}(2) \) (Option b)
- \( \alpha = (4n + 2)\pi - \tan^{-1}(2) \) (Option d)