if : `p*cos theta=sqrt(p^(2)+q^(2)).sin theta, "then" : q * sin theta=`

Options

A. p
B. `sqrt(p^(2)-q^(2))`
C. `sqrt(p^(2)+q^(2)). cos theta`
D. none of theses

Answer

To solve the equation \( p \cos \theta = \sqrt{p^2 + q^2} \sin \theta \) and find the value of \( q \sin \theta \), we can follow these steps:

### Step 1: Square both sides of the equation
Starting with the given equation:
\[ p \cos \theta = \sqrt{p^2 + q^2} \sin \theta \]
We square both sides to eliminate the square root:
\[ (p \cos \theta)^2 = (\sqrt{p^2 + q^2} \sin \theta)^2 \]
This simplifies to:
\[ p^2 \cos^2 \theta = (p^2 + q^2) \sin^2 \theta \]

### Step 2: Expand the right side
Expanding the right side gives us:
\[ p^2 \cos^2 \theta = p^2 \sin^2 \theta + q^2 \sin^2 \theta \]

### Step 3: Rearrange the equation
Rearranging the equation to isolate terms involving \( q^2 \):
\[ p^2 \cos^2 \theta - p^2 \sin^2 \theta = q^2 \sin^2 \theta \]
Factoring out \( p^2 \) on the left side:
\[ p^2 (\cos^2 \theta - \sin^2 \theta) = q^2 \sin^2 \theta \]

### Step 4: Use the identity for cosine of double angle
We can use the trigonometric identity \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \):
\[ p^2 \cos 2\theta = q^2 \sin^2 \theta \]

### Step 5: Solve for \( q \sin \theta \)
To find \( q \sin \theta \), we rearrange the equation:
\[ q^2 \sin^2 \theta = p^2 \cos 2\theta \]
Taking the square root of both sides:
\[ q \sin \theta = \sqrt{p^2 \cos 2\theta} \]
This simplifies to:
\[ q \sin \theta = p \sqrt{\cos 2\theta} \]

### Final Answer
Thus, the value of \( q \sin \theta \) is:
\[ q \sin \theta = p \sqrt{\cos 2\theta} \]

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