Đáp án:
`a) (sin^3\alpha+cos^3\alpha)/(sin\alpha+cos\alpha)`
`=((sin\alpha+cos\alpha)(sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha))/(sin\alpha+cos\alpha)`
`=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha`
`=(sin^2\alpha+cos^2\alpha)-sin\alpha.cos\alpha`
`=1-sin\alpha.cos\alpha (\text{đpcm})`
`b) sin^4\alpha+cos^4\alpha-sin^6\alpha-cos^6\alpha`
`=(sin^2\alpha+cos^2\alpha)^2-2sin^2\alpha.cos^2\alpha-(sin^6\alpha+cos^6\alpha)`
`=1-2sin^2\alpha.cos^2\alpha-(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha.cos^2\alpha+cos^4\alpha)`
`=1-sin^2\alpha.cos^2\alpha-[(sin^2\alpha+cos^2\alpha)^2-3sin^2\alpha.cos^2\alpha]`
`=1-sin^2\alpha.cos^2\alpha-(1-3sin^2\alpha.cos^2\alpha)`
`=1-sin^2\alpha.cos^2\alpha-1+3sin^2\alpha.cos^2\alpha`
`=sin^2\alpha.cos^2\alpha (\text{đpcm})`
`c) (tan\alpha-tan\beta)/(cot\beta-cot\alpha)`
$=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}$
$=\dfrac{\dfrac{sin\alpha.cos\beta-sin\beta.cos\alpha}{cos\alpha.cos\beta}}{\dfrac{cos\beta.sin\alpha-cos\alpha.sin\beta}{sin\alpha.sin\beta}}$
`=(sin\alpha.sin\beta)/(cos\alpha.cos\beta)`
`=(sin\alpha)/(cos\alpha).(sin\beta)/(cos\beta)`
`=tan\alpha.tan\beta (\text{đpcm})`
`d) cos^4\alpha-sin^4\alpha`
`=(cos^2\alpha-sin^2\alpha)(cos^2\alpha+sin^2\alpha)`
`=cos^2\alpha-sin^2\alpha`
`=2cos^2\alpha-(sin^2\alpha+cos^2\alpha)`
`=2cos^2\alpha-1 (\text{đpcm})`
Giải thích các bước giải:
Ta có các tính chất
`\qquad sin^2\alpha+cos^2\alpha=1`
`\qquad tan\alpha=(sin\alpha)/(cos\alpha)`
`\qquad cot\alpha=(cos\alpha)/(sin\alpha)`
