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