From the following figure, find the values of : (i) cos B (ii) tan C (iii) sin^2B + cos^2B (iv) sin B.cos C + cos B.sin C

In Δ BAC, ⇒ BC2 = AB2 \+ AC2 (∵ BC is hypotenuse) ⇒ (17)2 = (8)2 \+ AC2 ⇒ 289 = 64 \+ AC2 ⇒ AC2 = 289 \- 64 ⇒ AC2 = 225 ⇒ AC = ⇒ AC = 15 (i) cos Hence, cos . (ii) tan Hence, tan . (iii) sin2B + cos2B Hence, sin2B + cos2B = 1. (iv) sin B.cos C + cos B.sin C Hence, sin B.cos C + cos B.sin C = 1.