Question #aa2dc

2 Answers
Mar 15, 2017

See proof below

Explanation:

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#

Mar 15, 2017

Yes, It does equal to tanx

Explanation:

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.