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Given that tan (A + B) = 1 and that tan (A-B) = 1/7, find without using tables the values oftan A and tan B |
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Answer» Tan (a + b) = 1 => tan (a - b) = 1/7 tan 2 a = tan (a+b + a -b) = [ tan(a+b) + tan (a-b) ] / [1 - tan (a+b) tan (a-b) ] = (1 + 1/7) / (1 - 1/7) = 4/3tan 2a = 2 tan a / (1 - tan² a) = 4/3 => 2 tan² a + 3 tan a - 2 = 0=> tan a = [-3 + - √[ (9 + 16) ] / 4 = -2 or 1/2 Now we have tan (a+b) = [tan a + tan b] / [ 1 - tan a tan b ]If tan a = -2, then => 1 = [ -2 + tan b ] / [ 1 + 2 tan b ] => tan b = -3 if tan a = 1/2 then, => 1 = [ 1 /2 + tan b ] / [ 1 - 1/2 tan b ] => tan b = 1/3 Tan (a + b) = 1 => tan (a - b) = 1/7 tan 2 a = tan (a+b + a -b) = [ tan(a+b) + tan (a-b) ] / [1 - tan (a+b) tan (a-b) ] = (1 + 1/7) / (1 - 1/7) = 4/3tan 2a = 2 tan a / (1 - tan² a) = 4/3 => 2 tan² a + 3 tan a - 2 = 0=> tan a = [-3 + - √[ (9 + 16) ] / 4 = -2 or 1/2 Now we have tan (a+b) = [tan a + tan b] / [ 1 - tan a tan b ]If tan a = -2, then => 1 = [ -2 + tan b ] / [ 1 + 2 tan b ] => tan b = -3 if tan a = 1/2 then, => 1 = [ 1 /2 + tan b ] / [ 1 - 1/2 tan b ] => tan b = 1/3 |
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