cos210∘+cos250∘−cos10∘cos50∘
Using, cos2x=2cos2x−1, ⇒cos2x=21+cos2x, we get
=21[1+cos20∘+1+cos100∘−2cos50∘cos10∘]
=21[2+cos100∘+cos20∘−cos60∘−cos40∘]
=21[2+cos100∘+cos20∘−21−cos40∘]
Using, cosC+cosD=2cos(2C+D)cos(2C−D), we get
=21[23+2cos60∘cos40∘−cos40∘]
=21[23+2×21×cos40∘−cos40∘]
=21[23+cos40∘−cos40∘]
=21×23=43.