Question

If $3\sin \theta +5\cos \theta =5$ then show that $5\sin \theta -3\cos \theta =\pm 3$

Answer 1

given $3\sin \theta +5\cos \theta =5$

square both the sides, we get,

$(3\sin \theta +5\cos \theta {)}^2=5^2$ use $(a+b{)}^2=a^2+b^2+2ab$

$\Rightarrow 9{\sin }^2\theta +25{\cos }^2\theta +\mathrm{2.3.5.}\sin \theta .\cos \theta =25$

using ${\sin }^2\theta +{\cos }^2\theta =1$, we get,

$\Rightarrow 9(1-{\cos }^2\theta )+25(1-{\sin }^2\theta )+30\sin \theta \cos \theta =25$

$\Rightarrow 9+25-9{\cos }^2\theta -25{\sin }^2\theta +30\sin \theta \cos \theta =25$

$\Rightarrow 9=9{\cos }^2\theta +25{\sin }^2\theta -30\sin \theta \cos \theta$

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

$(\because (a-b{)}^2=(b-a{)}^2)$

If $a^2=b^2$ then $a=\pm 6$

So $(5\sin \theta -3\cos \theta {)}^2=3^2$

$\Rightarrow (5\sin \theta -3\cos \theta )=\pm 3$

Hence Proved.