Question

Question #aa2dc

Answer 1

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`

Answer 2

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.