Given that sin theta + 2cos theta =1, then prove that 2sin theta - cos theta =2
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If sinA+2cosA=1then by squaring both sides..
we get 4cos^2A+sin^2A+4cosAsinA=1
..then 4(1-sin^2A)+1-cos^2A+4cosAsinA=1
..so ..4-4sin^2A+1-cos^2A+4cosAsinA=1.....-4sin^2A+4cosAsinA-cos^2A=-4..transposing sign..we get. ..
4sin^2A-4cosAsinA+cos^2A=4...(2sinA-cosA)^2=4..
2sinA-cosA=2...hence proved..
we get 4cos^2A+sin^2A+4cosAsinA=1
..then 4(1-sin^2A)+1-cos^2A+4cosAsinA=1
..so ..4-4sin^2A+1-cos^2A+4cosAsinA=1.....-4sin^2A+4cosAsinA-cos^2A=-4..transposing sign..we get. ..
4sin^2A-4cosAsinA+cos^2A=4...(2sinA-cosA)^2=4..
2sinA-cosA=2...hence proved..
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