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

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Posted on

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)

Posted on

First, we'll write the tangent identity:

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

We'll write the formulas for the sine and cosine of the sum of angles a and b:

sin(a+b) = sina*cosb + sinb*cosa

cos(a+b) = cosa*cosb - sina*sinb

We'll substitute sin(a+b) and cos(a+b) by their formulas:

tan(a+b) = (sina*cosb + sinb*cosa)/(cosa*cosb - sina*sinb)

We'll factorize by cosa*cosb:

tan(a+b) =cosa*cosb*[(sina*cosb/cosa*cosb) + (sinb*cosa/cosa*cosb)]/cosa*cosb*[1 - (sina*sinb/cosa*cosb)]

We'll simplify and we'll get:

tan(a+b) = (sina/cos a + sinb/cos b)/(1 - tan a*tan b)

tan(a+b) = (tan a + tan b)/(1 - tan a*tan b) q.e.d.

Posted on

L:H:S = 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)

devide the numerator and denominator by cos a.cos b

= (sin a/cos a + sin b/cos b) / (1-sin a.sin b/cos a.cos b)

= (tan a + tan b)/(1 - tan a.tan b)

= R:H:S

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