(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)
