Question

If $3\sin \theta +4\cos \theta =5$, then the value of $4\sin \theta -3\cos \theta$ is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

$3\sin \theta +4\cos \theta =5$
Squaring both sides, we get
$9{\sin }^2\theta +16{\cos }^2\theta +24\sin \theta \cos \theta =25\Rightarrow 9(1-{\cos }^2\theta )+16(1-{\sin }^2\theta )+24\sin \theta \cos \theta =25\Rightarrow 25-9{\cos }^2\theta -16{\sin }^2\theta +24\sin \theta \cos \theta =25\Rightarrow 9{\cos }^2\theta +16{\sin }^2\theta -24\sin \theta \cos \theta =0\Rightarrow (4\sin \theta -3\cos \theta {)}^2=0\therefore 4\sin \theta -3\cos \theta =0$



Alternate method:
$3\sin \theta +4\cos \theta =5$
Let $4\sin \theta -3\cos \theta =x$
Squaring and adding both the equation, we get
$9+16=25+x^2\Rightarrow x=0\therefore 4\sin \theta -3\cos \theta =0$