(i) Prove sin(θ)/cos(θ) + cos(θ)/sin(θ) = 2cosec(2θ) , (ii) draw draph of y = 2cosec(2θ) for 0<θ< 360°, (iii) solve to 1 d.p. : sin(θ)/cos(θ) + cos(θ)/sin(θ) = 3.

For (i) Start by looking at left hand side of the equation. It is usually much easier to reduce large expression into smaller ones. So, LHS = sin²(θ) +cos²(θ)/ sin(θ)cos(θ). Since, sin²(θ) +cos²(θ) = 1 and the fact that sin(2θ) = 2sin(θ)cos(θ) we can reduce our expression into ⇒ LHS = 2 × [1 / sin(2θ) ]. We know, resiprocal of sin(θ) is cosec(θ), so simplying further we get 2cosec(2θ) = RHS.For (ii) first draw a sin(2θ) graph on side. Now, for y = 2cosec(2θ) check if there are any assymptotes , this is where the points would tend to infinity. Check maximum and mimimum points. Then using factor 2 which means our y values are stretched so for each x, new y value is double its old value. For(iii) using part (i) we can say 2cosec(2θ) = 3 ⇒ sin(2θ) = 2/3, find a new interval to solve this equation 0< 2θ <720°. Use CAST(Cos All Sin Tan) trick to find all possible values to (2 d.p. at this stage). All possible values for 2θ = sin^-1(2/3) = 41.81° , 138.19°, 401.81°,498.19° ⇒ θ = 2θ /2 = 20.9°, 69.1°, 200.9°, 249.1° (to 1 d.p.)