Giả sử `ΔABC` vuông tại `A`
a) Gọi `\alpha` là $\widehat{C}$ ta có:
`\tan\alpha=(\sin\alpha)/(\cos\alpha)`
`\tanC=(\sinC)/(\cosC)`
`\tanC=\sinC:\cosC`
`\tanC=(AB)/(BC):(AC)/(BC)`
`\tanC=(AB)/(BC).(BC)/(AC)`
`\tanC=(AB)/(AC)`
Vậy `\tan\alpha=(\sin\alpha)/(\cos\alpha)` `(đpcm)`
Gọi `\alpha` là $\widehat{C}$ ta có:
`\tan\alpha=(\cos\alpha)/(\sin\alpha)`
`\cotC=(\cosC)/(\sinC)`
`\cotC=\cosC:\sinC`
`\cotC=(AC)/(BC):(AB)/(BC)`
`\cotC=(AC)/(BC).(BC)/(AB)`
`\cotC=(AC)/(AB)`
Vậy `\cot\alpha=(\cos\alpha)/(\sin\alpha)` `(đpcm)`
b) Gọi `alpha` là $\widehat{B}$ ta có:
`\tan\alpha.\cot\alpha`
`=\tanB.\cotB`
`=(AC)/(AB).(AB)/(AC)`
`=1`
Vậy `\tan\alpha.\cot\alpha=1` `(đpcm)`
c) Gọi `\alpha` là $\widehat{B}$ ta có:
`\sin^2\alpha+\cos^2\alpha`
`=\sin^2B+\cos^2B`
`=((AC)/(BC))^2+((AB)/(BC))^2`
`=(AC^2)/(BC^2)+(AB^2)/(BC^2)`
`=(AC^2+AB^2)/(BC^2)`
`=(BC^2)/(BC^2)`
`=1`
Vậy `\sin^2\alpha+\cos^2\alpha=1` `(đpcm)`
