Question

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

• A. 16
• B. 25
• C. 9
• D. 50

Answer 1

The correct option is B 25

Given:
$acos\theta +bsin\theta =4$ and $asin\theta -bcos\theta =3$

Squaring on both the sides we get,
$(acos\theta +bsin\theta {)}^2=4^2\Rightarrow a^{2}co{s}^2\theta +b^{2}si{n}^2\theta +2absin\theta cos\theta =16....(i)$

$(asin\theta -bcos\theta {)}^2=3^2\Rightarrow a^{2}si{n}^2\theta +b^{2}co{s}^2\theta -2absin\theta cos\theta =\mathrm{9.....}(ii)$

Adding, (i) and (ii), we get
$a^{2}co{s}^2\theta +b^{2}si{n}^2\theta +2absin\theta cos\theta +a^{2}si{n}^2\theta +b^{2}co{s}^2\theta -2absin\theta cos\theta =16+9$

On simplifying the above terms, we get,
$a^2(co{s}^2\theta +si{n}^2\theta )+b^2(co{s}^2\theta +si{n}^2\theta )=25$

Since $si{n}^2\theta +co{s}^2\theta =1$ we get,
$a^2\left(1\right)+b^2\left(1\right)$ = 25

$\Rightarrow a^2+b^2=25$