Question

Show that (sinΘ+cosΘ)²+ (sinΘ-cosΘ)²=2

Answer 1

$(sin\theta+cos\theta)^2+(sin\theta-cos\theta)^2\\ \\ =[sin^2\theta+cos^2\theta+2\ sin\theta\ cos\theta]+[sin^2\theta+cos^2\theta-2\ sin\theta\ cos\theta]\\ \\ =sin^2\theta+sin^2\theta+cos^2\theta+cos^2\theta+2\ sin\theta\ cos\theta-2\ sin\theta\ cos\theta\\ \\=2sin^2\theta+2cos^2\theta\\ \\ =2(sin^2\theta+cos^2\theta)\\ \\=2 \times 1\\ \\=2$