Assuming you meant #cotx *cosy + siny#.
#color(blue)("Method 1":#
Let #color(Red)(sigma = cotxcosy+siny#. We know that #cotx = cosx/sinx#, so we have:
#sigma = (cosxcosy)/sin x + siny#
Let's get rid of the #sin x# denominator, by multiplying both sides by it.
#sigmacolor(blue)(sinx)=cosxcosy+sinxsiny#
Notice how
#cosxcosy+sinxsiny=cos(x-y)#.
#:. sigmacolor(blue)sinx = cos(x-y)#
But we know that #x-y = pi/2#, so we have
#sigmacolor(blue)sinx=cos(pi/2)=0#
From this, we get that either #sigma# or #sinx# is equal to zero.
However, only #color(red)("one")# of these solutions is true.
If #sinx =0#, then #cotx# is undetermined, because we would have to divide by #0#.
Thus,
#color(red)(sigma=0)#
#color(red)(cotxcosy+siny=0#, if #x-y=pi/2#
#color(blue)("Method 2:"#
If #x-y = pi/2#, then #x=pi/2+y#. Substituting it into our equation, we have
#cot(pi/2+y)cosy+siny = cos(pi/2+y)/sin(pi/2+y) * cosy + siny#
We have to use the sum formulas for trigonometric functions:
#cos(a+b)=cosacosb-sinasinb#
#sin(a+b)=sinacosb+cosasinb#
#:. cos(pi/2+y)=cos(pi/2)cosy-sin(pi/2)siny#
#= 0 *cosy-1*siny=-siny#
#:. sin(pi/2+y)=sin(pi/2)cosy+cos(pi/2)siny#
#=1*cosy+0*siny=cosy#
Thus, we get
#cot(pi/2+y)cosy+siny=-siny/cosy * cosy+siny = 0#
#color(red)(cotxcosy+siny=0#, if #x-y=pi/2#
