if x sin theta = y cos theta = 2z tan theta / (1-tan^theta) then find 4z2(x2 + y^2) in terms of x and y.

Given the equation:

\[
x \sin \theta = y \cos \theta = \frac{2z \tan \theta}{1 - \tan^2 \theta}
\]

We are tasked with finding \(4z^2 (x^2 + y^2)\) in terms of \(x\) and \(y\).

### Step 1: Set a common variable
Let us set the common value as \(k\), so:

\[
x \sin \theta = k, \quad y \cos \theta = k, \quad \frac{2z \tan \theta}{1 - \tan^2 \theta} = k
\]

### Step 2: Solve for \(\sin \theta\) and \(\cos \theta\)
From the first two equations, we get:

\[
\sin \theta = \frac{k}{x}, \quad \cos \theta = \frac{k}{y}
\]

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute the values of \(\sin \theta\) and \(\cos \theta\):

\[
\left( \frac{k}{x} \right)^2 + \left( \frac{k}{y} \right)^2 = 1
\]

This simplifies to:

\[
\frac{k^2}{x^2} + \frac{k^2}{y^2} = 1
\]

Multiplying through by \(x^2 y^2\):

\[
k^2 (y^2 + x^2) = x^2 y^2
\]

Solving for \(k^2\):

\[
k^2 = \frac{x^2 y^2}{x^2 + y^2}
\]

### Step 3: Solve for \(z\)
Now, use the third equation:

\[
\frac{2z \tan \theta}{1 - \tan^2 \theta} = k
\]

We know \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}\), so substitute this:

\[
\frac{2z \frac{y}{x}}{1 - \left( \frac{y}{x} \right)^2} = k
\]

Simplifying the denominator:

\[
1 - \frac{y^2}{x^2} = \frac{x^2 - y^2}{x^2}
\]

Thus, the equation becomes:

\[
\frac{2z \frac{y}{x}}{\frac{x^2 - y^2}{x^2}} = k
\]

Simplifying further:

\[
2z \frac{y}{x} \cdot \frac{x^2}{x^2 - y^2} = k
\]

\[
\frac{2z y x}{x^2 - y^2} = k
\]

Solving for \(z\):

\[
z = \frac{k (x^2 - y^2)}{2xy}
\]

### Step 4: Calculate \(4z^2 (x^2 + y^2)\)
We already know that:

\[
z = \frac{k (x^2 - y^2)}{2xy}
\]

So,

\[
z^2 = \frac{k^2 (x^2 - y^2)^2}{4x^2 y^2}
\]

Now, multiply by \(4(x^2 + y^2)\):

\[
4z^2 (x^2 + y^2) = 4 \cdot \frac{k^2 (x^2 - y^2)^2}{4x^2 y^2} \cdot (x^2 + y^2)
\]

\[
= \frac{k^2 (x^2 - y^2)^2 (x^2 + y^2)}{x^2 y^2}
\]

Substitute \(k^2 = \frac{x^2 y^2}{x^2 + y^2}\) into the equation:

\[
4z^2 (x^2 + y^2) = \frac{\frac{x^2 y^2}{x^2 + y^2} (x^2 - y^2)^2 (x^2 + y^2)}{x^2 y^2}
\]

Simplifying:

\[
4z^2 (x^2 + y^2) = (x^2 - y^2)^2
\]

### Final Answer:

\[
4z^2 (x^2 + y^2) = (x^2 - y^2)^2
\]

Would you like more details or have any questions?

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Here are five related questions to expand your understanding:
1. How does the Pythagorean identity help solve for trigonometric expressions in equations?
2. What is the significance of the variable \(k\) in simplifying the problem?
3. How does substitution of trigonometric ratios simplify complex expressions?
4. Can this method be applied to similar trigonometric identity problems?
5. How does solving for one variable influence the calculation of others in multi-variable equations?

**Tip:** Always start by simplifying trigonometric identities using known identities like the Pythagorean identity or tangent and cotangent definitions.

Learn how to solve a trigonometric equation with variables x, y, and z, using key trigonometric identities like the Pythagorean identity and algebraic manipulation. This detailed solution walks through each step, ideal for Grades 11-12 students.

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