Question

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

Answer 1

We have,

$3\sin \theta +4cos\theta =5$ ………. (1)

On squaring both sides, we get

${(3\sin \theta +4cos\theta )}^2=5^2$

$9{\sin }^2\theta +16{\cos }^2\theta +24\sin \theta \cos \theta =25$

$9(1-{\cos }^2\theta )+16(1-{\sin }^2\theta )+12\times 2\sin \theta \cos \theta =25$

$9-9{\cos }^2\theta +16-16{\sin }^2\theta +12\times 2\sin \theta \cos \theta =25$

$25-9{\cos }^2\theta -16{\sin }^2\theta +12\times 2\sin \theta \cos \theta =25$

$9{\cos }^2\theta +16{\sin }^2\theta -12\times 2\sin \theta \cos \theta =0$

${(3\cos \theta -4sin\theta )}^2=0$

$3\cos \theta -4sin\theta =0$

Hence, the value is $0$.