If CosA - SinA =m and CosA + SinA = n,show that m 2 - n 2 m 2 +n 2
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cosA-sinA=m and cosA+sinA=n
∴, m+n=cosA-sinA+cosA+sinA=2cosA
m-n=cosA-sinA-cosA-sinA=-2sinA
mn=(cosA-sinA)(cosA+sinA)=cos²A-sin²A
∴, m²-n²/m²+n²
=(m+n)(m-n)/{(m+n)²-2mn}
=2cosA(-2sinA)/{(2cosA)²-2(cos²A-sin²A)}
=(-4sinAcosA)/(4cos²A-2cos²A+2sin²A)
=(-4sinAcosA)/2(sin²A+cos²A) [∵, sin²A+cos²A=1]
=-2sinAcosA
∴, m+n=cosA-sinA+cosA+sinA=2cosA
m-n=cosA-sinA-cosA-sinA=-2sinA
mn=(cosA-sinA)(cosA+sinA)=cos²A-sin²A
∴, m²-n²/m²+n²
=(m+n)(m-n)/{(m+n)²-2mn}
=2cosA(-2sinA)/{(2cosA)²-2(cos²A-sin²A)}
=(-4sinAcosA)/(4cos²A-2cos²A+2sin²A)
=(-4sinAcosA)/2(sin²A+cos²A) [∵, sin²A+cos²A=1]
=-2sinAcosA
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