Question

If $4\sin \theta +3\cos \theta =5$, then find the value of ${\left(3\sin \theta -4\cos \theta \right)}^2$

Answer 1

We have,

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

On squaring both sides, we get

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

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

$16(1-{\cos }^2\theta )+9(1-{\sin }^2\theta )+24\sin \theta \cos \theta =25$

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

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

$-16{\cos }^2\theta -9{\sin }^2\theta +24\sin \theta \cos \theta =0$

$9{\sin }^2\theta +16{\cos }^2\theta -24\sin \theta \cos \theta =0$

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

Hence, the value is $0$.