Let M denote the determinant of a square matrix M . Let g : 0 , π 2 → ℝ be the function defined by g θ = f θ - 1 + f π 2 - θ - 1 where f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + sin π cos θ + π 4 tan θ - π 4 sin θ - π 4 - cos π 2 log e 4 π cot θ + π 4 log e π 4 tan π Let p x be a qua

Step-by-step solution

Given, f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + sin π cos θ + π 4 tan θ - π 4 sin θ - π 4 - cos π 2 log e 4 π cot θ + π 4 log e π 4 tanπ ⇒ f θ = 1 2 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 + 0 cos θ + π 4 tan θ - π 4 sin θ - π 4 0 log e 4 π - tan θ - π 4 - log e 4 π 0 Here we used cos θ + π 4 = - sin θ - π 4 And tan