Prove the following trigonometric identities.

Solution:

Given: $3 \sin \theta+5 \cos \theta=5$

We have to prove that $5 \sin \theta-3 \cos \theta=\pm 3$.

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Squaring the given equation, we have

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

$\Rightarrow \quad 9 \sin ^{2} \theta+2 \times 3 \sin \theta \times 5 \cos \theta+25 \cos ^{2} \theta=25$

$\Rightarrow 9\left(1-\cos ^{2} \theta\right)+2 \times 3 \sin \theta \times 5 \cos \theta+25\left(1-\sin ^{2} \theta\right)=25$

$\Rightarrow 9-9 \cos ^{2} \theta+2 \times 3 \sin \theta \times 5 \cos \theta+25-25 \sin ^{2} \theta=25$

$\Rightarrow \quad 34-\left(9 \cos ^{2} \theta-2 \times 3 \sin \theta \times 5 \cos \theta+25 \sin ^{2} \theta\right)=25$

$\Rightarrow \quad-\left(25 \sin ^{2} \theta-2 \times 5 \sin \theta \times 3 \cos \theta+9 \cos ^{2} \theta\right)=-9$

$\Rightarrow \quad\left(25 \sin ^{2} \theta-2 \times 5 \sin \theta \times 3 \cos \theta+9 \cos ^{2} \theta\right)=9$

$\Rightarrow \quad(5 \sin \theta-3 \cos \theta)^{2}=9$

$\Rightarrow \quad 5 \sin \theta-3 \cos \theta=\pm 3$

Hence proved.