Question : If $a\sin\theta+b\cos\theta=c$, then $a\cos\theta-b\sin\theta$ is equal to:
Option 1: $\pm \sqrt{a+b-c}$
Option 2: $\pm \sqrt{a^{2}+b^{2}+c^{2}}$
Option 3: $\pm \sqrt{a^{2}+b^{2}-c^{2}}$
Option 4: $\pm \sqrt{c^{2}+a^{2}-b^{2}}$
Correct Answer: $\pm \sqrt{a^{2}+b^{2}-c^{2}}$
Solution :
Given $a\sin\theta+b\cos\theta=c$
Squaring both sides, we get:
$(a\sin\theta+b\cos\theta)^2=c^2$
⇒ $a^2\sin^2\theta+b^2\cos^2\theta+2ab\sin\theta\cos\theta=c^2$
⇒ $a^2(1-\cos^2\theta)+b^2(1-\sin^2\theta)+2ab\sin\theta\cos\theta=c^2$
⇒ $a^2-a^2\cos^2\theta+b^2-b^2\sin^2\theta+2ab\sin\theta\cos\theta=c^2$
⇒ $-a^2\cos^2\theta-b^2\sin^2\theta+2ab\sin\theta\cos\theta=c^2-a^2-b^2$
⇒ $-(a^2\cos^2\theta+b^2\sin^2\theta-2ab\sin\theta\cos\theta)=-(-c^2+a^2+b^2)$
⇒ $(a\cos\theta-b\sin\theta)^2=(-c^2+a^2+b^2)$
⇒ $(a\cos\theta-b\sin\theta)=\pm\sqrt{a^2+b^2-c^2}$
Hence, the correct answer is $\pm\sqrt{a^2+b^2-c^2}$.
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