Question

If $a=\cos \theta +b\sin \theta =c$, then the value of $(a\sin \theta -b\cos \theta )$ is

• A. $\pm \sqrt{a^2+b^2-c^2}$
• B. $\pm \sqrt{a^2-b^2+c^2}$
• C. $\pm \sqrt{a^2+b^2+c^2}$
• D. $\pm \sqrt{b^2+c^2-a^2}$

Answer 1

Answer: (A)

$\pm \sqrt{a^2+b^2-c^2}$

$a\cos \theta +b\sin \theta =c$

Squaring both sides, we get

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

$\Rightarrow a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta \cos \theta =c^2$

$\Rightarrow a^2(1-{\sin }^2\theta )+b^2(1-{\cos }^2\theta )+2ab\sin \theta \cos \theta =c^2$

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

$\Rightarrow a^2+b^2-[a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta ]=c^2$

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

$\Rightarrow {(a\cos \theta -b\sin \theta )}^2=a^2+b^2-c^2$
$\Rightarrow (a\cos \theta -b\sin \theta )=\pm \sqrt{a^2+b^2-c^2}$

Hence, the answer is $\pm \sqrt{a^2+b^2-c^2}.$