Question

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

Answer 1

Given :
$3\sin \theta +5\cos \theta =5$
Squaring on both the sides we get,
$(3\sin \theta +5\cos \theta {)}^2=5^2$
$9{\sin }^2\theta +25{\cos }^2\theta +30\sin \theta \cos \theta =25$ $[\because (a+b)²=a²+b²+2ab]$
$9(1-{\cos }^2\theta )+25(1-{\sin }^2\theta )+30\sin \theta \cos \theta =25$ [$\because \sin ²\theta +\cos ²\theta =1$]
$9-9\cos ²\theta +25-25\sin ²\theta +30\sin \theta \cos \theta =25$
$-9\cos ²\theta -25\sin ²\theta +30\sin \theta \cos \theta =25-34$
$-9\cos ²\theta -25\sin ²\theta +30\sin \theta \cos \theta =-9$
$9\cos ²\theta +25\sin ²\theta -30\sin \theta \cos \theta =9$
$25\sin ²\theta +9\cos ²\theta -30\sin \theta \cos \theta =9$
$(5\sin \theta -3\cos \theta )²=9$ $\because \left[\right(a-b)²=a²+b²-2ab]$
$(5\sin \theta -3\cos \theta )=\surd 9$
$(5\sin \theta -3\cos \theta )=\pm 3$