Question

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

• A. $-5$
• B. $-2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Given,

$3\sin \left(\theta \right)+4\cos \left(\theta \right)=5$

$3\sin \left(\theta \right)=5-4\cos \left(\theta \right)$

$\Rightarrow {\left(3\sin \left(\theta \right)\right)}^2={(5-4\cos \left(\theta \right))}^2$

$9{\sin }^2\left(\theta \right)-{(5-4\cos \left(\theta \right))}^2=0$

$-{(5-4\cos \left(\theta \right))}^2+(1-{\cos }^2\left(\theta \right))\cdot 9=0$

$-16-25{\cos }^2\left(\theta \right)+40\cos \left(\theta \right)=0$

let, $\cos \left(\theta \right)=u$

$-16-25u^2+40u=0$

$-25u^2+40u-16=0$

$u_{1,2}=\frac{-40\pm \sqrt{0}}{2(-25)}\to u=\frac{-40}{2(-25)}$

$\Rightarrow u=\frac{4}{5}$

$\therefore \cos \left(\theta \right)=\frac{4}{5}$

$\Rightarrow \sin \left(\theta \right)=\frac{3}{5}$

Now,

$4\sin \left(\theta \right)-3\cos \left(\theta \right)$

$=4\times \frac{3}{5}-3\times \frac{4}{5}$

$=\frac{12}{5}-\frac{12}{5}$

$=0$