Question

If acos theta + bsin theta =c ,then prove that asin theta -bcos theta =+- square root under a2 +b2-c2

Answer 1

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Answer 2

$a\sin \theta-b\cos \theta$=±$\sqrt{a^{2}+b^{2}-c^{2}}$, proved.

Step-by-step explanation:

We have,

$a\cos \theta+b\sin \theta=c$...... (1)

To prove, $a\sin \theta-b\cos \theta$=±$\sqrt{a^{2}+b^{2}-c^{2} }$  

Let $a\sin \theta-b\cos \theta=x$     ...... (2)

Squaring and adding (1) and (2), we get

$(a^{2} \cos ^{2} \theta+b^{2} \sin^{2} \theta+2ab\sin \theta\sin \theta)+(a^{2} \sin ^{2} \theta+b^{2} \cos^{2} \theta-2ab\sin \theta\sin \theta)=c^{2} +x^{2}$

⇒$(a^{2} \cos ^{2} \theta+b^{2} \sin^{2} \theta)+(a^{2} \sin ^{2} \theta+b^{2} \cos^{2} \theta)=c^{2} +x^{2}$

⇒ $a^{2}( \cos ^{2} \theta+\sin^{2} \theta)+b^{2} (\sin ^{2} \theta+ \cos^{2} \theta)=c^{2} +x^{2}$

⇒ $a^{2}(1)+b^{2} (1)=c^{2} +x^{2}$

⇒ $x^{2}=a^{2}+b^{2}-c^{2}$

⇒x = ± $\sqrt{a^{2}+b^{2}-c^{2}}$, proved.

Hence, $a\sin \theta-b\cos \theta$]=±$\sqrt{a^{2}+b^{2}-c^{2}}$.