Question : If $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$, then the value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –2
Correct Answer: 2
Solution :
Given:
$x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$
So, $\frac{x}{a}=(\sin\theta+\cos\theta), \frac{y}{a}=(\sin\theta-\cos\theta)$
Now, $\frac{x^2}{a^2}+\frac{y^2}{b^2}$
= $(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2$
= $\sin^2\theta+2×\sin\theta×\cos\theta+\cos^2\theta+\sin^2\theta-2×\sin\theta×\cos\theta+\cos^2\theta$
= $\sin^2\theta+\cos^2\theta+\sin^2\theta+\cos^2\theta$
= $1+1$
= $2$
Hence, the correct answer is $2$.
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