Question : If $x=a\left ( \sin\theta+\cos\theta \right )$ and $y=b\left ( \sin\theta-\cos\theta \right )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:
Option 1: 4
Option 2: 3
Option 3: 1
Option 4: 2
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Correct Answer: 2
Solution :
Given: $x=a\left ( \sin\theta+\cos\theta \right )$ and $y=b\left ( \sin\theta-\cos\theta \right )$.
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = \frac{a^{2}\left ( \sin^{2}\theta+2\sin\theta\cos\theta+\cos^{2}\theta \right )}{a^{2}} + \frac{b^{2}\left ( \sin^{2}\theta-2\sin\theta\cos\theta+\cos^{2}\theta \right )}{b^{2}}$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = \left ( \sin^{2}\theta+2\sin\theta\cos\theta+\cos^{2}\theta \right ) + \left ( \sin^{2}\theta-2\sin\theta\cos\theta+\cos^{2}\theta \right )$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 2\sin^{2}\theta+2\cos^{2}\theta$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 2(\sin^{2}\theta+\cos^{2}\theta)$
Since $\sin^{2}\theta+\cos^{2}\theta=1$,
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$
Hence, the correct answer is 2.
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