Question

Question #82a02

Answer 1

Here X= `Tan^-1((1-(3)^(1/2)]/(1+(3)^(1/2)))`

Explanation:

There is an identity `Tan(x+y)` = `(Tan(x)+tan(y))/(1-tan(x)tan(y)`

When we apply it to your equation we get
`tan(x+60)=(tan(x)+tan(60))/(1-tan(x)tan(60)`
As we know `tan(60)=(3)^(1/2)`

Therefore `(tan(x)+(3)^(1/2))/(1-(3)^(1/2)tanx)=1`

`tan(x)+(3)^(1/2)=1-(3)^(1/2)tan(x)`

`tan(x)+(3)^(1/2)tan(x)=1-(3)^(1/2)`

`tan(x)(1+(3)^(1/2))=1-(3)^(1/2)`

`tan(x)=(1-(3)^(1/2))/(1+(3)^(1/2)`

Therefore `x=tan^-1((1-(3)^(1/2))/(1+(3)^(1/2)))`

Answer 2

A different approach...
See the answer below...

Explanation:

`tan(x+60)=1`
`=>tan(x+60)=tan45`(mainly)

We might take further higher value...
`x+60=45`
`=>x=-15`

Hope this helps...

It is same as `tan^-1((1-sqrt3)/(1+sqrt3))`...