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What is \(\frac{sin^6\,θ-cos^6\,θ}{sin^2\,θ-cos^2\,θ}\) equal to?(a) sin4θ – cos4θ (b) 1 – sin2θ cos2θ (c) 1 + sin2θ cos2θ (d) 1 – 3 sin2θ cos2θ |
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Answer» (b) 1 – sin2 θ cos2 θ \(\frac{sin^6\,θ-cos^6\,θ}{sin^2\,θ-cos^2\,θ}\) = \(\frac{(sin^2\,\theta)^3-(cos^2\,\theta)^3}{sin^2\,\theta-cos^2\,\theta}\) = \(\frac{(sin^\,\theta-cos^2\,\theta)(sin^4\,\theta+cos^4\,\theta+sin^2\,\theta\,\cos^2\,\theta)}{(sin^2\,\theta-cos^2\,\theta)}\) (Using a3 – b3 = (a – b) (a2 + b2 + ab)) = sin4 θ + cos4 θ + 2 sin2 θ cos2 θ – sin2 θ cos2 θ = (sin2 θ + cos2 θ)2 – sin2 θ cos2 θ = 1 – sin2 θ cos2 θ (∵ sin2 θ + cos2 θ = 1) |
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