Use the property #color(blue)(cos(A+B)=cosAcosB-sinAsinB#
For the #color(red)(tan^-1 (-1)# we know that our triangle is in quadrant 4. Since #color(red)(tan^-1 x# is restricted to quadrants 1 and 4 and since the argument is negative it means our triangle has to be in quadrant 4.
Hence, the side #color(red)(opposite # to #color(red)(angle theta = -1# and #color(red)(adjacent# to #color(red)(angle theta = 1# and so the #color(red)(hypote n use=sqrt2#. In this case #color(red)(theta=45^@# but we don't need to know the value of #color(red)(theta#. Therefore,
#color(blue)(cos(tan^-1(-1))=cos theta=(adjacent)/(hypote n use)=1/sqrt2 = sqrt2/2#
#color(blue)(sin(tan^-1(-1))=sin theta=(opposite)/(hypote n use)=-1/sqrt2 = -sqrt2/2#
For the #color(magenta)(cos^-1(-4/5)#the triangle is in quadrant 2 since #color(magenta)(cos^-1 x)#is restricted to quadrants 1 and 2 and since the argument is negative the triangle has to be in quadrant 2.
Hence, #color(magenta)(adjacent # to #color(magenta)(angle theta=-4# and #color(magenta)(hypote n use = 5#. Therefore by using pythagorean theorem the #color(magenta)(opposite# to #color(magenta)(angle theta=3#
#color(blue)((cos(cos^-1(-4/5))=cos theta = (adjacent)/(hypote n use)=-4/5#
#color(blue)(sin(cos^-1(-4/5)=sin theta=(opposite)/(hypote n use) = 3/5#
Then using the property #color(orange)(cos(A+B)=cosAcosB-sinAsinB# where #color(orange)(A=tan^-1 (-1) and B=cos^-1(-4/5)# we have
#cos(tan^-1(-1)+cos^-1(-4/5))=color(blue)(cos(tan^-1(-1))cos(cos^-1(-4/5))-sin(tan^-1(-1)) sin(cos^-1(-4/5))#
#color(blue)(=sqrt2/2 * (-4)/5-(-sqrt2)/2*3/5#
#color(blue)(=(-4sqrt2)/10)+color(blue)((3sqrt2)/10#
#color(blue)( :. = -sqrt2/10#
