if 2cos=root 3 find value of 2cos Q+1/2cosQ-1..urgently tomorrow is my exam
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2cosQ=√3
or, cosQ=√3/2
∴, 2cosQ+1/2cosQ-1
={(2×√3/2)+1}/{(2×√3/2)-1}
=(√3+1)/(√3-1)
=(√3+1)(√3+1)/(√3-1)(√3+1)
=(√3+1)²/{(√3)²-(1)²}
=(3+2√3+1)/(3-1)
=(4+2√3)/2
=2+√3
or, cosQ=√3/2
∴, 2cosQ+1/2cosQ-1
={(2×√3/2)+1}/{(2×√3/2)-1}
=(√3+1)/(√3-1)
=(√3+1)(√3+1)/(√3-1)(√3+1)
=(√3+1)²/{(√3)²-(1)²}
=(3+2√3+1)/(3-1)
=(4+2√3)/2
=2+√3
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