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

Solution

1. Let $y=4\sin (x)-3\cos (x)$ and consider the given equation $3\sin (x)+4\cos (x)=5$.
2. Square both equations to get $(3\sin (x)+4\cos (x){)}^2=25$ and $y^2=(4\sin (x)-3\cos (x){)}^2$.
3. Expand both squares to obtain $9{\sin }^2(x)+16{\cos }^2(x)+24\sin (x)\cos (x)=25$ and $y^2=16{\sin }^2(x)+9{\cos }^2(x)-24\sin (x)\cos (x)$.
4. Add the two squared equations together to get $25{\sin }^2(x)+25{\cos }^2(x)=25+y^2$.
5. Apply the identity ${\sin }^2(x)+{\cos }^2(x)=1$ to simplify the equation to $25=25+y^2$, which implies $y^2=0$.
6. Hence the answer is (A).