Question

If $acos\theta +bsin\theta =4$ and $asin\theta -bcos\theta =3$, then $a^2+b^2=$___.

• A. 7
• B. 12
• C. 25
• D. None of these

Answer 1

The correct option is C 25

$acos\theta +bsin\theta =4$ and $asin\theta -bcos\theta =3$
Squaring both the equations,
$(acos\theta +bsin\theta {)}^2=4^2\text{and}(asin\theta -bcos\theta {)}^2=3^2\text{Adding, we get}{a}^{2}co{s}^2\theta +2absin\theta cos\theta +b^{2}si{n}^2\theta +a^{2}si{n}^2\theta -2absin\theta cos\theta +b^{2}co{s}^2\theta =16+9\Rightarrow a^2(co{s}^2\theta +si{n}^2\theta )+b^2(co{s}^2\theta +si{n}^2\theta )=25\Rightarrow a^2+b^2=25$