Question : If $x=a\cos\theta + b\sin\theta$ and $y=b\cos\theta-a\sin\theta$, then $x^{2}+y^{2}$ is equal to:
Option 1: $ab$
Option 2: $a^2+b^2$
Option 3: $a^2-b^2$
Option 4: $1$
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Correct Answer: $a^2+b^2$
Solution :
Given: $x=a\cos\theta + b\sin\theta$ and $y=b\cos\theta-a\sin\theta$
To find: $x^{2}+y^{2}$
Putting the value of $x$ and $y$, we get
$(a\cos\theta+b\sin\theta)^{2}+(b\cos\theta-a\sin\theta)^{2}$
$(a^2\cos^2\theta+b^2\sin^2\theta+2×a\cos\theta×b\sin\theta)+(b^2\cos^2\theta+a^2\sin^2\theta-2×b\cos\theta×a\sin\theta)$
$=a^2\cos^2\theta + b^2\sin^2\theta+b^2\cos^2\theta+a^2\sin^2\theta$
$=a^2(\cos^2\theta+sin^2\theta)+b^2(sin^2\theta+\cos^2\theta)$
We know that, $(sin^2\theta+\cos^2\theta)=1$
Thus, $a^2(1)+b^2(1)$= $a^2+b^2$
Hence, the correct answer is $a^2+b^2$.
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