cos theta/cosec theta+1+cos theta /cosec theta-1=2 tan theta Share with your friends Share 9 Manbar Singh answered this LHS = cos θcosec θ + 1 + cos θcosec θ - 1 = cos θ cosec θ - 1 + cos θ cosec θ + 1cosec θ + 1cosec θ - 1 = cos θ . cosec θ - cos θ + cos θ cosec θ + cos θcosec2θ - 1 = 2 cos θ cosec θcot2θ = 2 cos θ × cosec θ × 1cot2θ = 2 cos θ × 1sin θ × sin2θcos2θ = 2 × sin θcos θ = 2 tan θ = RHS 35 View Full Answer