Suppose that x=2t/(1+t²) and y=(1-t²)/(1+t²).Show that the value of x²+y² does not depend on the value of t.

Answer:x² + y² = 1Step-by-step explanation:Hi there!We two functions of tx = 2t/(1 + t²)y = (1 - t²) / (1 + t²)First, let´s square x:x² = [2t/(1 + t²)]²Apply distributive propertyx² = (2t)²/(1+t²)²x² = 4t²/(1+t²)(1+t²)Apply distributive propertyx² = 4t² /(1 + 2t² + t⁴)Now let´s square y:y² = [(1 - t²) / (1 + t²)]²Apply distributive propertyy² = (1 - t²)² / (1 + t²)²y² = (1 - t²)(1 - t²) / (1 + t²)(1 + t²)Apply distributive propertyy² = (1 - 2t² + t⁴) / ( 1 + 2t² + t⁴)x² + y² = [4t² /(1 + 2t² + t⁴)] + [(1 - 2t² + t⁴) / ( 1 + 2t² + t⁴)]Since the denominators are equal, we have to sum the numeratorsx² + y² = (4t² + 1 - 2t² + t⁴)/ (1 + 2t² + t⁴)x² + y² = (1 + 2t² + t⁴) / (1 + 2t² + t⁴)x² + y² = 1Then, x² + y² = 1  and therefore does not depend on the value of t