Question

tính \(\frac{2}{\tan\alpha-1}+\frac{\cos\alpha+1}{\cot\alpha-1}\)

b) \(2\left(\sin^6\alpha+\cos^6\alpha\right)-3\left(\sin^4\alpha+\cos^4\alpha\right)\)

Answer 1

Câu a chắc bạn ghi nhầm \(\frac{cota+1}{cota-1}\) thành \(\frac{cosa+1}{cota-1}\)

\(\frac{2}{tana-1}+\frac{cota+1}{cota-1}=\frac{2cota}{1-cota}+\frac{cota+1}{cota-1}=\frac{-2cota+cota+1}{cota-1}=\frac{1-cota}{-\left(1-cota\right)}=-1\)

\(2\left(sin^6x+cos^6x\right)-3\left(sin^4x+cos^4x\right)\)

\(=2\left(sin^2x+cos^2x\right)^3-6sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3\left(sin^2x+cos^2x\right)^2+6sin^2x.cos^2x\)

\(=-1-6sin^2x.cos^2x+6sin^2x.cos^2x=-1\)