If `2ycostheta=x sintheta` and `2xsectheta-ycosectheta=3`, then `x^(2)+4y^(2)=?`

Options

A. 4
B. `-4`
C. `pm4`
D. 0

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Substitute the value of θ
We will choose θ = 45° (or π/4 radians) to simplify our calculations.

### Step 2: Calculate the values of sin(θ) and cos(θ)
Using θ = 45°, we have:
- sin(45°) = 1/√2
- cos(45°) = 1/√2

### Step 3: Substitute into the first equation
The first equation is:
\[ 2y \cos(\theta) = x \sin(\theta) \]
Substituting the values of sin and cos:
\[ 2y \cdot \frac{1}{\sqrt{2}} = x \cdot \frac{1}{\sqrt{2}} \]
This simplifies to:
\[ 2y = x \]
Thus, we have:
\[ x = 2y \] (Equation 1)

### Step 4: Substitute into the second equation
The second equation is:
\[ 2x \sec(\theta) - y \csc(\theta) = 3 \]
Using sec(θ) = 1/cos(θ) and csc(θ) = 1/sin(θ):
\[ 2x \cdot \sqrt{2} - y \cdot \sqrt{2} = 3 \]
Substituting Equation 1 (x = 2y):
\[ 2(2y) \cdot \sqrt{2} - y \cdot \sqrt{2} = 3 \]
This simplifies to:
\[ 4y\sqrt{2} - y\sqrt{2} = 3 \]
Combining like terms:
\[ 3y\sqrt{2} = 3 \]

### Step 5: Solve for y
Dividing both sides by 3:
\[ y\sqrt{2} = 1 \]
Thus:
\[ y = \frac{1}{\sqrt{2}} \]

### Step 6: Substitute back to find x
Using Equation 1 (x = 2y):
\[ x = 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 7: Calculate x² + 4y²
Now we need to find:
\[ x^2 + 4y^2 \]
Substituting the values of x and y:
\[ x^2 = (\sqrt{2})^2 = 2 \]
\[ 4y^2 = 4 \left(\frac{1}{\sqrt{2}}\right)^2 = 4 \cdot \frac{1}{2} = 2 \]
Thus:
\[ x^2 + 4y^2 = 2 + 2 = 4 \]

### Final Answer
The value of \( x^2 + 4y^2 \) is \( 4 \).

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