Assuming, {1-(sinx)^2}/cosx
{1-(sinx)^2}/cosx
or {(cosx)^2}/cosx , since 1-(sinx)^2 = (cosx)^2
or cosx -----------(1)
Given, tanx =1 so perpendicular = 1, base=1 so hypotenuse = sqrt( Perpendicular^2 + base^2) = sqrt(1+1) = sqrt(2)
So, from (1) we know,
cosx = base/hypotenuse
cosx = 1/sqrt(2)
1/sqrt(2) is the solution