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Prove the following identities:If $ 3 \sin \theta+5 \cos \theta=5 $, prove that $ 5 \sin \theta-3 \cos \theta=\pm 3 $.
Given:
\( 3 \sin \theta+5 \cos \theta=5 \)
To do:
We have to prove that \( 5 \sin \theta-3 \cos \theta=\pm 3 \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
Therefore,
$3 \sin \theta+5 \cos \theta=5$
Squaring on both sides, we get,
$(3 \sin \theta+5 \cos \theta)^2=(5)^2$
$\Rightarrow 9 \sin ^{2} \theta+25 \cos ^{2} \theta+30 \sin \theta \cos \theta=25$ $\Rightarrow 9\left(1-\cos ^{2} \theta\right)+25\left(1-\sin ^{2} \theta\right)+30 \sin \theta \cos \theta=25$
$\Rightarrow 9-9 \cos ^{2} \theta+25-25 \sin ^{2} \theta+30 \sin \theta \cos \theta=25$
$\Rightarrow -25 \sin ^{2} \theta-9 \cos ^{2} \theta+30 \sin \theta \cos \theta=25-9-25$
$\Rightarrow -25 \sin ^{2} \theta-9 \cos ^{2} \theta+30 \sin \theta \cos \theta=-9$
$\Rightarrow 25 \sin ^{2} \theta+9 \cos ^{2} \theta-30 \sin \theta \cos \theta=9$
$\Rightarrow (5 \sin \theta)^{2}+(3 \cos \theta)^{2}-2 \times 5 \sin \theta \times 3 \cos \theta=(\pm 3)^{2}$
$\Rightarrow (5 \sin \theta-3 \cos \theta)^{2}=(\pm 3)^{2}$
$\Rightarrow 5 \sin \theta-3 \cos \theta=\pm 3$
Hence proved.