tan(a+b)=3^1/2,tan(a-b)=1 then tan10a=
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tan (a+b) = √3 => a+b = π/3
tan (a-b) = 1 => a-b = π/4
2a = 7 π / 12
10 a = 35 π/12 = 2 π + 11 π/12
tan 10 a = tan (11 π/12) = tan (π - π/12)
= - tan π/12
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another way:
tan 2a = tan (a+b + a -b) = [tan (a+b) + tan (a-b) ] / [ 1 - tan (a+b) tan (a-b) ]
= [ √3 + 1 ] / [ 1 - √3 ] = - (√3+1)² /2 = - (2 +√3)
tan 4a = 2 tan 2a / [ 1 - tan² 2a ] = - 2 (2 +√3)/ [ 1 - 4 - 3 - 4√3 ]
= (2 +√3) / [√3 (2 + √3) ] = 1/√3
tan 8a = (2/√3) / [ 1 - 1/3 ] = √3
tan 10 a = tan (8a +2a) = [ √3 - 2 -√3 ] / [ 1 + √3(2+√3) ]
= - 1 / (2 + √3) = - (2 - √3)
tan (a-b) = 1 => a-b = π/4
2a = 7 π / 12
10 a = 35 π/12 = 2 π + 11 π/12
tan 10 a = tan (11 π/12) = tan (π - π/12)
= - tan π/12
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another way:
tan 2a = tan (a+b + a -b) = [tan (a+b) + tan (a-b) ] / [ 1 - tan (a+b) tan (a-b) ]
= [ √3 + 1 ] / [ 1 - √3 ] = - (√3+1)² /2 = - (2 +√3)
tan 4a = 2 tan 2a / [ 1 - tan² 2a ] = - 2 (2 +√3)/ [ 1 - 4 - 3 - 4√3 ]
= (2 +√3) / [√3 (2 + √3) ] = 1/√3
tan 8a = (2/√3) / [ 1 - 1/3 ] = √3
tan 10 a = tan (8a +2a) = [ √3 - 2 -√3 ] / [ 1 + √3(2+√3) ]
= - 1 / (2 + √3) = - (2 - √3)
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