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  • Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

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Solution.  Given:-   sinθ + 2cosθ = 1
squaring both sides we have
\left ( \sin \theta +2\cos \theta \right )^{2}= 1^{2}
\sin ^{2}\theta +4\cos ^{2}\theta +4\sin \theta \cos \theta = 1
\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )
\sin ^{2}\theta +4\cos ^{2}\theta = 1-4\sin \theta \cos \theta \cdots \left ( 1 \right )
To prove :

2\sin \theta -\cos \theta = 2
Taking the left-hand side
2\sin \theta -\cos \theta \cdots \left ( 2 \right )
On squaring equation (2) we get
\left (2 \sin \theta -\cos \theta \right )^{2}
= 4\sin ^{2}\theta +\cos ^{2}\theta -4\sin \theta \cos \theta  \left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )
= 3\sin ^{2}\theta +\sin ^{2}\theta+\cos ^{2}\theta -4\sin \theta \cos \theta
= 3\sin ^{2}\theta +1-4\sin \theta \cos \theta         \left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]
= 3\sin ^{2} \theta+\sin^{2} \theta+4\cos ^{2} \theta           [use equation (1)]
= 4\sin ^{2}\theta +4\cos ^{2}\theta
= 4\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )
= 4 

 \left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]
So here we get the value of (2sin\theta – cos\theta)2 is 4
\left ( 2\sin \theta -\cos \theta \right )^{2}= 4
2\sin \theta -\cos \theta = \sqrt{4}
2\sin \theta -\cos \theta = 2
Hence proved

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