Question

cos 2θ cos 2ϕ + sin²(θ – ϕ) – sin²(θ + ϕ) is equal to
(a) sin 2 (θ + ϕ)
(b) cos 2 (θ + ϕ)
(c) sin 2 (θ – ϕ)
(d) cos 2 (θ – ϕ)

Answer 1

$$\begin{gathered} \cos 2\theta \cos 2\varphi +{\sin }^2\left(\theta -\varphi \right)-{\sin }^2\left(\theta +\varphi \right) \\ =\cos 2\theta \cos 2\varphi +\sin \left(\theta -\varphi +\theta +\varphi \right)\sin \left(\theta -\varphi -\theta -\varphi \right)\left(\mathrm{using}\mathrm{identity}{\sin }^{2}A-{\sin }^{2}B=\sin \left(A+B\right)\sin \left(A-B\right)\right) \\ =\cos 2\theta \cos 2\varphi +\sin \left(2\theta \right)\sin \left(-2\varphi \right) \\ =\cos 2\theta \cos 2\varphi -\sin 2\theta \sin 2\varphi \left(\because \sin \left(-x\right)=-\sin x\right) \\ =\cos \left(2\theta +2\varphi \right)\left(\mathrm{using}\mathrm{identity}:\cos a\cos b-\sin a\sin b=\cos \left(a+b\right)\right) \\ =\cos 2\left(\theta +\varphi \right) \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{B}. \end{gathered}$$