Show that 1−cos2x1+cos2x≡tan2x\displaystyle \frac{1 - \cos 2x}{1 + \cos 2x} \equiv \tan^2 x1+cos2x1−cos2x≡tan2x
Hence solve, for −π≤θ≤π-\pi \leq \theta \leq \pi−π≤θ≤π,
1−cos2x1+cos2x=3 \frac{1 - \cos 2x}{1 + \cos 2x} = 3 1+cos2x1−cos2x=3137 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.