The expression cos4θ−sin4θ+2sin2θ+3(cosecθ+cotθ+1)(cosecθ−cotθ+1),00<θ<900\frac{cos^4\theta-sin^4\theta + 2 sin^2 \theta+3 }{(cosec \theta + cot \theta +1)(cosec \theta - cot \theta+1)}, 0^0<\theta<90^0(cosecθ+cotθ+1)(cosecθ−cotθ+1)cos4θ−sin4θ+2sin2θ+3,00<θ<900 is equal to: A12sinθ\frac{1}{2}sin\theta21sinθ B2 sin θ\thetaθ Csec θ\thetaθ D2 cosec θ\thetaθ Show answer & explanationCorrect answer: BExplanationThe correct answer is option 2 i.e. 2 sin θ\thetaθ cos4θ−sin4θ+2sin2θ+3(cosecθ+cotθ+1)(cosecθ−cotθ+1),00<θ<900\frac{cos^4\theta-sin^4\theta + 2 sin^2 \theta+3 }{(cosec \theta + cot \theta +1)(cosec \theta - cot \theta+1)}, 0^0<\theta<90^0(cosecθ+cotθ+1)(cosecθ−cotθ+1)cos4θ−sin4θ+2sin2θ+3,00<θ<900 = (cos2θ)2−(sin2θ)2+2sin2θ+3(cosecθ+1+cotθ)(cosecθ+1−cotθ)\frac{(cos^2\theta)^2-(sin^2\theta)^2 + 2 sin^2 \theta+3 }{(cosec \theta + 1 + cot \theta) (cosec \theta + 1 - cot \theta)}(cosecθ+1+cotθ)(cosecθ+1−cotθ)(cos2θ)2−(sin2θ)2+2sin2θ+3 = 1+3(cosecθ+1)2−(cotθ)2\frac{1+3 }{(cosec \theta + 1)^2 - (cot \theta)^2}(cosecθ+1)2−(cotθ)21+3 = 4(cosec2θ+1+2cosecθ)−(cot2θ)\frac{4}{(cosec^2 \theta + 1 + 2cosec \theta)- (cot^2 \theta)}(cosec2θ+1+2cosecθ)−(cot2θ)4 = (4)/(2 cosecθ\thetaθ) = 2 sinθ\thetaθ.