Question #aa2dc

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Answer 1 · 2017-03-15T07:26:14

See proof below

We use

$sin(a-b)=sinacosb-sinbcosa$

$cos(a-b)=cosacosb-sinasinb$

So,

$tan(a-b)=sin(a-b)/cos(a-b)$

$=(sinacosb-sinbcosa)/(cosacosb-sinasinb)$

Dividing by $cosacosb$

$tan(a-b)=((sinacosb)/(cosacosb)-(sinbcosa)/(cosacosb))/((cosacosb)/(cosacosb)-(sinasinb)/(cosacosb))$

$=(tana-tanb)/(1-tanatanb)$

Let $a=x+y$

and $b=y$

$tan(a-b)=tan(x+y-y)=tanx=RHS$

$(tan(x+y)-tany)/(1-tan(x+y)tany)=LHS$

So,

$(tan(x+y)-tany)/(1-tan(x+y)tany)=tanx$

$QED$

Answer 2 · 2017-03-15T07:30:39

Yes, It does equal to tanx

let's assume that $(x+y) = a and y = b$

When we substitute it in the equation,
We get,

$(tana-tanb)/(1+tanatanb)$ ....... (1)

And we know that,

$tan(A-B) = (tanA-tanB)/(1+tanAtanB)$

So, we can conclude that,

$(tana-tanb)/(1+tanatanb) = tan(a-b)$

As we know that,

$(x+y) = a and y = b$

So by substituting values we get,
$tan(a-b) => tan(x+y-y) => tanx$

Hence Proved.