Question

If a $cos\theta +bsin\theta =3$ & $asin\theta -bcos\theta =4$ , then $a^2+b^2$ has the value =

• A. 25
• B. 14
• C. 7
• D. 10

Answer 1

The correct option is A 25

Given,

$a\cos \theta +b\sin \theta =3$

$a\sin \theta +b\cos \theta =4$

squaring adding both the above equation, we get,

$a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\cos }^2\theta +{\sin }^2\theta )+2ab\cos \theta \sin \theta -2ab\cos \theta \sin \theta =3^3+4^2$

$a^2\left(1\right)+b^{(}1)=9+16$

$a^2+b^2=25$