Let α and, β be such that  π<α−β<3π. If sin⁡α+sin⁡β=2165 and cos⁡α+cos⁡β=−2765, then the value of cos⁡α−β2 is

We have  sin⁡α+sin⁡β=2165-----icos⁡α+cos⁡β=−2765----iiSquaring Eq. (i), we getsin2⁡α+sin2⁡β+2sin⁡αsin⁡β=21652---iiiSquaring Eq. (ii), we getcos2⁡α+cos2⁡β+2cos⁡αcos⁡β=27652----ivAdding Eqs. (iii) and (iv), we get2+2cos⁡(α−β)=1(65)2(27)2+(21)2=1(65)2(729+441)=1(65)2(1170)=1865or  1+cos⁡(α−β)=965or  2cos2⁡α−β2=965or cos⁡α−β2=−3130∵π<α−β<3π ⇒π2<α−β2<3π2⇒cos⁡α−β2<0