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