Show that
cos2xcosx−sin2xsinx≡−secx,x≠nπ2, n∈Z \frac{\cos 2x}{\cos x} - \frac{\sin 2x}{\sin x} \equiv -\sec x, \quad x \neq \frac{n\pi}{2}, \; n \in \mathbb{Z} cosxcos2x−sinxsin2x≡−secx,x=2nπ,n∈ZHence solve, for 0<θ<π0 < \theta < \pi0<θ<π,
(cos2θcosθ−sin2θsinθ)2=5−4tanθ \left( \frac{\cos 2\theta}{\cos \theta} - \frac{\sin 2\theta}{\sin \theta} \right)^2 = 5 - 4\tan \theta (cosθcos2θ−sinθsin2θ)2=5−4tanθgiving your answers to 3 significant figures as appropriate.
Using the result from part (a), or otherwise, find the exact value of
∫0π3(cos2xcosx−sin2xsinx)tanx dx \int_{0}^{\frac{\pi}{3}} \left( \frac{\cos 2x}{\cos x} - \frac{\sin 2x}{\sin x} \right) \tan x \, dx ∫03π(cosxcos2x−sinxsin2x)tanxdx
