1.

cos 2θ cos 2ϕ + sin2(θ–ϕ) – sin2(θ + ϕ) is equal toA. sin 2(θ + ϕ)B. cos 2(θ + ϕ)C. sin 2(θ–ϕ)D. cos 2(θ–ϕ)[Hint: Use sin2A – sin2B = sin (A + B) sin (A – B)]

Answer»

B. cos 2(θ + ϕ)

Given that, cos 2θ cos 2ϕ + sin2(θ–ϕ) – sin2(θ + ϕ)

= cos 2θ cos 2ϕ + sin(θ–ϕ+ θ + ϕ)sin(θ–ϕ- θ – ϕ)

[∵sin2A – sin2B = sin (A + B) sin (A – B)]

= cos 2θ cos 2ϕ + sin 2θ. sin(- 2ϕ)

= cos 2θ cos 2ϕ - sin 2θ. Sin 2ϕ [∵ sin(-θ) = -sin θ]

= cos 2(θ + ϕ) [∵ cos x cos y – sin x sin y = cos(x + y)]



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