If cos(A+B) = 0 and cot (A - B) =root 3, then evaluate (i) cosA. cosB - sinA. sinB (ii) cot A - cot B cotA. cotB+1 Share with your friends Share 5 Varun.Rawat answered this We have, cosA+B = 0⇒cosA+B = cos 90°⇒A + B = 90° ......1 cotA - B = 3⇒cotA - B = cot 30°⇒A - B = 30° .......2Adding 1 and 2, we get2A = 120°⇒A = 60°From 1, we get B = 30°1 cos A . cos B - sin A . sin B=cos 60 . cos 30 - sin 60 . sin 30=12×32 - 32 × 12=02. cot A - cot Bcot A . cot B + 1=cot 60° - cot 30°cot 60° cot 30° + 1=13 - 313×3 + 1=1 - 323 = -13 7 View Full Answer Rachitajain answered this The value of A is 60and B is 30.i.) placing the values the answer comes 0ii.) again, substituting the values, the answer comes -1/ root 3 1