Show that 1tanθ−tanθ≡cos2θsinθcosθ\frac{1}{\tan \theta} - \tan \theta \equiv \frac{\cos 2\theta}{\sin \theta \cos \theta}tanθ1−tanθ≡sinθcosθcos2θ for θ≠nπ2\theta \neq \frac{n\pi}{2}θ=2nπ where n∈Zn \in \mathbb{Z}n∈Z.
Solve, for 0∘≤x<90∘0^\circ \le x < 90^\circ0∘≤x<90∘, the equation 5sin2(2x−15∘)=25 \sin^2(2x - 15^\circ) = 25sin2(2x−15∘)=2 giving your answers in degrees to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)
Practise Edexcel A Level Maths 7.5 Proving Trigonometric Identities with exam-style questions for A Level Maths. 27 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.