To solve for [tex]\( 4z^2(x^2 + y^2) \)[/tex] given the equation [tex]\( x \sin \theta = y \cos \theta = \frac{2z \tan \theta}{1 - \tan^2 \theta} \)[/tex], we should follow these steps:
1. Express [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex] and [tex]\( \theta \)[/tex]:
From [tex]\( x \sin \theta = \frac{2z \tan \theta}{1 - \tan^2 \theta} \)[/tex], we get:
[tex]\[
x = \frac{2z \tan \theta}{\sin \theta (1 - \tan^2 \theta)}
\][/tex]
Similarly, from [tex]\( y \cos \theta = \frac{2z \tan \theta}{1 - \tan^2 \theta} \)[/tex], we get:
[tex]\[
y = \frac{2z \tan \theta}{\cos \theta (1 - \tan^2 \theta)}
\][/tex]
2. Solve for [tex]\( 4z^2(x^2 + y^2) \)[/tex]:
Substitute the expressions of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] into [tex]\( 4z^2(x^2 + y^2) \)[/tex]:
[tex]\[
4z^2 \left( \left( \frac{2z \tan \theta}{\sin \theta (1 - \tan^2 \theta)} \right)^2 + \left( \frac{2z \tan \theta}{\cos \theta (1 - \tan^2 \theta)} \right)^2 \right)
\][/tex]
Simplifying each term inside the parentheses:
[tex]\[
\left( \frac{2z \tan \theta}{\sin \theta (1 - \tan^2 \theta)} \right)^2 = \frac{4z^2 \tan^2 \theta}{\sin^2 \theta (1 - \tan^2 \theta)^2}
\][/tex]
[tex]\[
\left( \frac{2z \tan \theta}{\cos \theta (1 - \tan^2 \theta)} \right)^2 = \frac{4z^2 \tan^2 \theta}{\cos^2 \theta (1 - \tan^2 \theta)^2}
\][/tex]
Adding these two terms:
[tex]\[
4z^2 \left( \frac{4z^2 \tan^2 \theta}{\sin^2 \theta (1 - \tan^2 \theta)^2} + \frac{4z^2 \tan^2 \theta}{\cos^2 \theta (1 - \tan^2 \theta)^2} \right)
\][/tex]
[tex]\[
= 4z^2 \times 4z^2 \tan^2 \theta \left( \frac{1}{\sin^2 \theta (1 - \tan^2 \theta)^2} + \frac{1}{\cos^2 \theta (1 - \tan^2 \theta)^2} \right)
\][/tex]
[tex]\[
= 16z^4 \tan^2 \theta \left( \frac{1}{\sin^2 \theta (1 - \tan^2 \theta)^2} + \frac{1}{\cos^2 \theta (1 - \tan^2 \theta)^2} \right)
\][/tex]
Now, observe that:
[tex]\[
\frac{1}{\sin^2 \theta (1 - \tan^2 \theta)^2} + \frac{1}{\cos^2 \theta (1 - \tan^2 \theta)^2} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta \cos^2 \theta (1 - \tan^2 \theta)^2}
\][/tex]
Since [tex]\( \sin^2 \theta + \cos^2 \theta = 1 \)[/tex]:
[tex]\[
\frac{1}{\sin^2 \theta \cos^2 \theta (1 - \tan^2 \theta)^2}
\][/tex]
Which simplifies to:
[tex]\[
16z^4 \tan^2 \theta \left( \frac{1}{\sin^2 \theta \cos^2 \theta (1 - \tan^2 \theta)^2} \right) = \frac{16z^4 \tan^2 \theta}{\sin^2 \theta \cos^2 \theta (1 - \tan^2 \theta)^2}
\][/tex]
3. Simplified expression:
Simplified form:
[tex]\[
4z^2(x^2 + y^2) = 32 \frac{z^4}{1 + \cos(4\theta)}
\][/tex]
Thus, the final result is:
[tex]\[ 32z^4 / (cos(4\theta) + 1) \][/tex]
This matches option (4):
[tex]\[ 4z^2(x^2 + y^2) = \boxed{\left( x^2 + y^2 \right)^2} \][/tex]
