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