If `abs(cos theta{sin theta+sqrt(sin^2theta+sin^2alpha)})lek`, then the value of k

Options

A. `sqrt(1+cos^2alpha)`
B. `sqrt(1+sin^2alpha)`
C. `sqrt(2+sin^2alpha)`
D. `sqrt(2+cos^2alpha)`

Explanation

To solve the problem, we need to find the value of \( k \) such that:

\[ |\cos \theta (\sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha})| \leq k \]

### Step 1: Define the expression
Let:
\[ u = \cos \theta (\sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha}) \]

### Step 2: Rearranging the expression
We can rearrange the expression:
\[ u - \sin \theta \cos \theta = \cos \theta \sqrt{\sin^2 \theta + \sin^2 \alpha} \]

### Step 3: Square both sides
Square both sides to eliminate the square root:
\[ (u - \sin \theta \cos \theta)^2 = \cos^2 \theta (\sin^2 \theta + \sin^2 \alpha) \]

### Step 4: Expand both sides
Expanding the left side:
\[ u^2 - 2u \sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta = \cos^2 \theta (\sin^2 \theta + \sin^2 \alpha) \]

### Step 5: Simplify the equation
Now, simplify the equation:
\[ u^2 - 2u \sin \theta \cos \theta = \cos^2 \theta \sin^2 \alpha \]

### Step 6: Rearranging the quadratic equation
Rearranging gives us:
\[ u^2 - 2u \sin \theta \cos \theta - \cos^2 \theta \sin^2 \alpha = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = 1 \)
- \( b = -2 \sin \theta \cos \theta \)
- \( c = -\cos^2 \theta \sin^2 \alpha \)

The discriminant \( D \) must be non-negative for real solutions:
\[ D = (-2 \sin \theta \cos \theta)^2 - 4(1)(-\cos^2 \theta \sin^2 \alpha) \geq 0 \]

### Step 8: Calculate the discriminant
Calculating the discriminant:
\[ D = 4 \sin^2 \theta \cos^2 \theta + 4 \cos^2 \theta \sin^2 \alpha \]
\[ D = 4 \cos^2 \theta (\sin^2 \theta + \sin^2 \alpha) \geq 0 \]

### Step 9: Find the maximum value of \( u \)
From the quadratic equation, we find that:
\[ u = \sin \theta \cos \theta \pm \sqrt{\cos^2 \theta (\sin^2 \theta + \sin^2 \alpha)} \]
To find the maximum value of \( |u| \), we consider:
\[ |u| \leq \sqrt{1 + \sin^2 \alpha} \]

### Step 10: Conclusion
Thus, we conclude that:
\[ k = \sqrt{1 + \sin^2 \alpha} \]

### Final Answer
The value of \( k \) is:
\[ \boxed{\sqrt{1 + \sin^2 \alpha}} \]