The value of sec - 1 ⁡ 1 4 ∑ k = 0 10 sec ⁡ 7 π 12 + k π 2 sec ⁡ 7 π 12 + k + 1 π 2 in the interval - π 4 , 3 π 4 equals

Step-by-step solution

sec − 1 1 4 ∑ k = 0 10 sec 7 π 12 + k π 2 sec 7 π 12 + k π 2 + π 2 = sec − 1 − 1 4 ∑ k = 0 10 sec 7 π 12 + k π 2 cosec 7 π 12 + k π 2 ∴ sec π 2 + θ = − cosec θ = s e c - 1 - 1 4 ∑ k = 0 10 1 cos ⁡ 7 π 12 + k π 2 sin ⁡ 7 π 12 + k π 2 = sec - 1 ⁡ - 1 4 ∑ k = 0 10 2 sin ⁡ 7 π 6 + k π ∴ 2 sin ⁡ θ cos ⁡ θ = sin ⁡ 2 θ