Question

Question 11
If $asin\theta +bcos\theta =c$, then prove that $acos\theta -bsin\theta =\sqrt{a^2+b^2-c^2}$

Answer 1

Given that, $asin\theta +bcos\theta =c$

On squaring both sides,

$(asin\theta +bcos\theta {)}^2=c^2$

$\Rightarrow a^{2}si{n}^2\theta +b^{2}co{s}^2\theta +2absin\theta .cos\theta =c^2$

$[\because (x+y{)}^2=x^2+2xy+y^2]$

$\Rightarrow a^2(1-co{s}^2\theta )+b^2(1-si{n}^2\theta )+2absin\theta .cos\theta =c^2$

$[\because si{n}^2\theta +co{s}^2\theta =1]$

$\Rightarrow a^2-a^{2}co{s}^2\theta +b^2-b^{2}si{n}^2\theta +2absin\theta .cos\theta =c^2$

$\Rightarrow a^2+b^2-c^2=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta -2absin\theta .cos\theta$

$\Rightarrow (a^2+b^2-c^2)=(acos\theta -bsin\theta {)}^2[\because a^2+b^2-2ab=(a-b{)}^2]$

$\Rightarrow (acos\theta -bsin\theta {)}^2=a^2+b^2-c^2$

$\Rightarrow acos\theta -bsin\theta =\sqrt{a^2+b^2-c^2}$