Derive tan(a+b), using sin (a +b) and cos(a +b).

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have sin (a+b) = (sin a * cos b + cos a * sin b) and cos(a + b) = (cos a * cos b – sin a * sin b)

Also 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 )]

cancel the common terms in the numerator and denominator and use sin x / cos x = tan x

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

Therefore we get:

tan (a +b) = (tan a + tan b)/(1 – tan a * tan b)

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neela | High School Teacher | (Level 3) Valedictorian

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To derive t(a+b) , using  sin(a+b) and cos(a+b).

We know the identities (i) sin(a+b) = sina*cosn+cosa*sinb and

(ii) cos(a+b) = csa*cos-sina*sinb.

Therefore tan(a+b) = sin(a+b)/cos(a+b).....(1).

We substitute sin(a+b) = sina*cosb+cos*sinb and cos(a+b) = cosa*cosb-sina*sin b in (1):

 tan (a+b) = (sina*cosb+cosa*sinb)/(cosa*cos-sina*sinb. We divide  both numerator and denominator in right by cosa*cosb and get:

tan(a+b) = {cosa*sinb/(cosa*cosb) - sina*cosb/(coa*cosb)}/{cosa*cosb/(cosacosb) - sina*sinb/(cosa*cosb)}

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

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