Column I Column II A In a triangle ∆ X Y Z let a , b and c be the lengths of the angles X , Y and Z , respectively. If 2 a 2 − b 2 = c 2 and λ = sin X − Y sinZ then possible values of n for which cos n π λ = 0 is are P 1 B In a triangle Δ X Y Z , let a , b and c be the lengths of the sides opposite to the angles X , Y and Z , respectively. If 1 + cos 2 x - 2 cos 2 y = 2 sin x sin

Step-by-step solution

A λ = sin ⁡ X - Y sin ⁡ Z sinX cosY - cosX sin Y sinZ a cosY - b cosX c a a 2 + c 2 - b 2 2 a c - b b 2 + c 2 - a 2 2 b c c = λ a 2 - b 2 c 2 = λ As 2 a 2 - b 2 = c 2 ∴ λ = 1 2 ∴ cos ⁡ n π 2 = 0 ∴ n = 1 , 3 , 5 B 1 + cos ⁡ 2 X - 2 cos ⁡ 2 Y = 2 sin ⁡ X sin ⁡ Y Solving 4 sin 2 Y - 2 sin 2 X = 2 s i n X s i n Y 4 b 2 - 2 a 2 = 2 a b 4 - 2 a b 2 = 2 a b a b = X X 2 + X - 2 = 0 X = 1 a b = 1 x = - 2 a b = - 2 (rejected) C O Y ≡ y = 3 x O X ≡ y = 1 3 x ∴ Equation of bisector of O X , O Y y = x ∴ x - y 2 = 3 2 β - 1 - β