prove  tan(a+b)=(tan a+tanb)/(1-tana*tanb)

Expert Answers

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Here, we use the relations that tan x = sin x / cos x.

sin (a + b) = sin a * cos b + cos a * sin b

cos ( a+ b) = cos a * cos b - sin a * sin b

tan (a + b) = sin (a + b) / cos ( a + b)

=> [sin a * cos b + cos a * sin b] / [cos a * cos b - sin a * sin b]

divide all the terms by cos a * cos b

=>[ (sin a * cos b)/(cos a * cos b)+ (cos a * sin b)/(cos a * cos b)] / [(cos a * cos b)/(cos a * cos b) - (sin a * sin b)/(cos a * cos b)]

=> [(sin a / cos a) + (sin b / cos b)]/[ 1 - (sin a / cos a)*( sin b/ cos b)]

=> (tan a + tan b) / ( 1 - tan a * tan b)

Therefore we have tan (a + b) = (tan a + tan b) / ( 1 - tan a * tan b)

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