If `2y cos theta =x sin theta and 2x sec theta -y cosec theta =3,` then `x ^(2) +4y^(2)=`

Options

A. 4
B. `-4`
C. `-2`
D. 2

Answer

To solve the problem, we have the following equations:

1. \( 2y \cos \theta = x \sin \theta \) (Equation 1)
2. \( 2x \sec \theta - y \csc \theta = 3 \) (Equation 2)

We need to find the value of \( x^2 + 4y^2 \).

### Step 1: Rearranging Equation 1

From Equation 1, we can express \( x \) in terms of \( y \):

\[ x = \frac{2y \cos \theta}{\sin \theta} \]

### Step 2: Substitute \( x \) in Equation 2

Now, substitute \( x \) into Equation 2:

\[ 2\left(\frac{2y \cos \theta}{\sin \theta}\right) \sec \theta - y \csc \theta = 3 \]

### Step 3: Simplifying the Equation

Recall that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \). Substitute these into the equation:

\[ 2\left(\frac{2y \cos \theta}{\sin \theta}\right) \cdot \frac{1}{\cos \theta} - y \cdot \frac{1}{\sin \theta} = 3 \]

This simplifies to:

\[ \frac{4y}{\sin \theta} - \frac{y}{\sin \theta} = 3 \]

Combining the terms gives:

\[ \frac{3y}{\sin \theta} = 3 \]

### Step 4: Solve for \( y \)

Multiplying both sides by \( \sin \theta \):

\[ 3y = 3 \sin \theta \]

Dividing by 3:

\[ y = \sin \theta \]

### Step 5: Substitute \( y \) back to find \( x \)

Now substitute \( y \) back into the expression for \( x \):

\[ x = \frac{2(\sin \theta) \cos \theta}{\sin \theta} = 2 \cos \theta \]

### Step 6: Calculate \( x^2 + 4y^2 \)

Now we can find \( x^2 + 4y^2 \):

\[ x^2 = (2 \cos \theta)^2 = 4 \cos^2 \theta \]

\[ 4y^2 = 4(\sin \theta)^2 = 4 \sin^2 \theta \]

Thus,

\[ x^2 + 4y^2 = 4 \cos^2 \theta + 4 \sin^2 \theta \]

Factoring out the 4:

\[ x^2 + 4y^2 = 4(\cos^2 \theta + \sin^2 \theta) \]

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ x^2 + 4y^2 = 4 \cdot 1 = 4 \]

### Final Answer

Thus, the value of \( x^2 + 4y^2 \) is:

\[ \boxed{4} \]