cos A minus cosB whole square minus sinA minus sinB whole square

cosA\[(cosA -cosB)^2+(sinA-sinB)^2=4sin^2((A-B)/2)\]

For A=180

      B=90

we get \[4sin^2((180-90)/2)\]

        =\[4sin^2(45)\]

       =2

Posted on by
Score
Views