You should remember that `tan x = sin x/cos x` and `tan y = sin y/cos y` , hence, substituting these fractions for `tan x` and `tan y` in the equation `tanx+tany=2` yields:

`sin x/cos x + sin y/cos y = 2`

You need to bring these fractions to a common denominator such that:

`(sin x*cos y + sin y*cos x)/(cos x*cos y) = 2 `

Notice that the problem provides the information that `2cosx*cosy=1` , hence, you may evaluate `cos x*cos y` such that:

`2cosx*cosy=1 => cosx*cosy=1/2`

Substituting `1/2` for `cosx*cosy in (sin x*cos y + sin y*cos x)/(cos x*cos y) = 2` yields:

`(sin x*cos y + sin y*cos x)/(1/2) = 2`

Notice that you may write `sin x*cos y + sin y*cos x` as `sin (x+y)` such that:

`2sin(x+y) = 2 => sin(x+y) = 1 => x+y = pi/2`

You may write `x = pi/2 - y` such that:

`tan(pi/2 - y) + tan y = 2`

`sin(pi/2 - y + y)/cos(pi/2-y)*cosy = 2`

`sin pi/2 = 2cos(pi/2-y)*cosy `

You should substitute `sin y` for `cos(pi/2-y)` such that:

`sin pi/2 = 2sin y*cos y`

You need to remember the equation that gives the sine of double angle such that:

`sin pi/2 = sin 2y =>2y = pi/2=> y = pi/4`

Since `x = pi/2 - y => x = pi/4`

**Hence, evaluating x and y under the given conditions yields `x = y = pi/4.` **

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