Question : If $r=15(\sin \theta+\cos \theta)$ and $s=16(\sin \theta-\cos \theta)$, then the value of $\frac{r^2}{15^2}+\frac{s^2}{16^2}$ is:
Option 1: 8
Option 2: 6
Option 3: 4
Option 4: 2
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Correct Answer: 2
Solution :
Given:
$r=15(\sin \theta+\cos \theta)$ and $s=16(\sin \theta-\cos \theta)$
$\frac{r^2}{15^2}+\frac{s^2}{16^2}$
$= \frac{15^2 (\sin\theta + \cos\theta)^2}{15^2} + \frac{16^2 (\sin\theta - \cos\theta)^2}{16^2}$
$= \sin^2\theta + \cos^2\theta + 2\sin\theta \cos\theta + \sin^2\theta + \cos^2\theta - 2\sin\theta \cos\theta$
$= 1 +2\sin\theta\cos\theta + 1 -2\sin\theta\cos\theta$
$= 2$
Hence, the correct answer is 2.
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