If `sin x + cosec x + tan y + cot y =4` where x and `y in [ 0 (pi)/(2)],` then `tan "" (y)/(2) ` is a root of the equation

Options

A. `alpha ^(2) + 2 alpha + 1=0`
B. `alpha ^(2) + 2 alpha -1 =0`
C. `2 alpha ^(2) - 2 alpha -1 =0`
D. `alpha ^(2) - alpha -1 =0`

Answer

To solve the equation \( \sin x + \csc x + \tan y + \cot y = 4 \) where \( x, y \in \left[0, \frac{\pi}{2}\right] \), we will analyze the components of the equation step by step.

### Step 1: Analyze the components
We know that:
- \( \sin x + \csc x \geq 2 \) (by AM-GM inequality)
- \( \tan y + \cot y \geq 2 \) (by AM-GM inequality)

### Step 2: Apply the inequalities
Using the inequalities from Step 1, we can combine them:
\[ \sin x + \csc x + \tan y + \cot y \geq 2 + 2 = 4 \]
This indicates that the equality \( \sin x + \csc x + \tan y + \cot y = 4 \) holds when both pairs achieve their minimum values.

### Step 3: Determine when equality holds
The equality in AM-GM holds when:
- \( \sin x = \csc x \) implies \( \sin x = 1 \) (thus \( x = \frac{\pi}{2} \))
- \( \tan y = \cot y \) implies \( \tan y = 1 \) (thus \( y = \frac{\pi}{4} \))

### Step 4: Substitute values
Substituting \( x = \frac{\pi}{2} \) and \( y = \frac{\pi}{4} \) into the original equation:
\[ \sin\left(\frac{\pi}{2}\right) + \csc\left(\frac{\pi}{2}\right) + \tan\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right) = 1 + 1 + 1 + 1 = 4 \]
This confirms that the equality holds.

### Step 5: Find \( \tan\left(\frac{y}{2}\right) \)
Now, we need to find \( \tan\left(\frac{y}{2}\right) \) where \( y = \frac{\pi}{4} \):
\[ \tan\left(\frac{y}{2}\right) = \tan\left(\frac{\pi/4}{2}\right) = \tan\left(\frac{\pi}{8}\right) \]

### Step 6: Set up the equation
Using the double angle identity for tangent:
\[ \tan y = \frac{2\tan\left(\frac{y}{2}\right)}{1 - \tan^2\left(\frac{y}{2}\right)} \]
Substituting \( \tan y = 1 \):
\[ 1 = \frac{2\tan\left(\frac{y}{2}\right)}{1 - \tan^2\left(\frac{y}{2}\right)} \]
Cross-multiplying gives:
\[ 1 - \tan^2\left(\frac{y}{2}\right) = 2\tan\left(\frac{y}{2}\right) \]
Rearranging leads to:
\[ \tan^2\left(\frac{y}{2}\right) + 2\tan\left(\frac{y}{2}\right) - 1 = 0 \]

### Step 7: Solve the quadratic equation
Let \( z = \tan\left(\frac{y}{2}\right) \):
\[ z^2 + 2z - 1 = 0 \]
Using the quadratic formula:
\[ z = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]
Since \( z = \tan\left(\frac{y}{2}\right) \) must be non-negative, we take:
\[ \tan\left(\frac{y}{2}\right) = -1 + \sqrt{2} \]

### Final Answer
Thus, \( \tan\left(\frac{y}{2}\right) \) is a root of the equation:
\[ z^2 + 2z - 1 = 0 \]