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:
Answer & explanation
Correct answer: option 4
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