If $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$, then the value of $\frac{x^2}{8^2}+\frac{y^2}{9^2}$ is:

2

Given :-

x = 8 ( sinθ + cosθ ) & y = 9 ( sinθ - cosθ )

\(\frac{x}{8}\) = ( sinθ + cosθ ) & \(\frac{y}{9}\) = ( sinθ - cosθ )  

Squaring on both side and on adding

\(\frac{x²}{8²}\) + \(\frac{y²}{9²}\) = ( sinθ + cosθ )² + ( sinθ - cosθ )² 

= sin²θ + cos²θ + 2 sinθ . cosθ + sin²θ + cos²θ  -  2 sinθ . cosθ 

= 1 + 1

= 2