Let S = θ ∈ 0 , 2 π : 8 2 sin 2 θ + 8 2 cos 2 θ = 16 . Then n S + ∑ θ ∈ S sec π 4 + 2 θ cosec π 4 + 2 θ is equal to:

Step-by-step solution

Given, 8 2 sin 2 θ + 8 2 cos 2 θ = 16 ⇒ 8 2 sin 2 θ + 8 2 - 2 sin 2 θ = 16 Now let 8 2 sin 2 θ = y ⇒ y + 64 y = 16 ⇒     y = 8 ⇒ 8 2 sin 2 θ = 8 ⇒ sin 2 θ = 1 2 ⇒ θ ∈ π 4 , 3 π 4 , 5 π 4 , 7 π 4 Now n S + ∑ θ ∈ S sec π 4 + 2 θ cosec π 4 + 2 θ = n S + ∑ θ ∈ S 1 cos π 4 + 2 θ sin π 4 + 2 θ = 4 + ∑ θ ∈ S 2 2 cos π