Question : 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:
Option 1: 4
Option 2: 6
Option 3: 8
Option 4: 2
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Correct Answer: 2
Solution :
Given: $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$
Now, $\frac{x^2}{8^2}+\frac{y^2}{9^2}$
= $\frac{[8(\sin \theta+\cos \theta)]^2}{8^2}+\frac{[9(\sin \theta-\cos \theta)]^2}{9^2}$
= $\frac{64(\sin^2 \theta+\cos^2 \theta+2\sin \theta\cos \theta)}{64}+\frac{81(\sin^2 \theta+\cos^2 \theta-2\sin \theta\cos \theta)}{81}$
= $1+2\sin \theta\cos \theta +1-2\sin \theta\cos \theta$
= $2$
Hence, the correct answer is 2.
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