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

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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 +...

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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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