Gọi cạnh đối là O, cạnh kề là A và cạnh huyền là H.
Ta có: \(\sin\partial=\frac{o}{H}\)
\(\cos\partial=\frac{A}{H}\)
\(\tan\partial=\frac{O}{A}\)
\(\cot\partial=\frac{A}{O}\)
a) \(\Rightarrow\frac{\sin\partial}{\cos\partial}=\frac{\frac{O}{H}}{\frac{A}{H}}=\frac{O}{A}=\tan\partial\)(đpcm)
b) \(\Rightarrow\frac{\cos\partial}{\sin\partial}=\frac{\frac{A}{H}}{\frac{O}{H}}=\frac{A}{O}=\cot\partial\)(đpcm)
c)\(\Rightarrow\tan\partial\times\cot\partial=\frac{O}{A}\times\frac{A}{O}=1\)(đpcm)
d)\(\Rightarrow\sin^2\partial+\cos^2\partial=\left(\frac{O}{H}\right)^2+\left(\frac{A}{H}\right)^2=\frac{O^2+A^2}{H^2}\)
Mà theo định lý Py-ta-go: \(H^2=O^2+A^2\)
\(\Rightarrow\frac{O^2+A^2}{H^2}=1\)\(\Leftrightarrow\sin^2\partial+\cos^2\partial=1\)(đpcm)
e) Ta có:\(1+\tan^2\partial=\)\(1+\left(\frac{O}{A}\right)^2=\frac{O^2+A^2}{A^2}=\frac{H^2}{A^2}\)(theo định lý Py-ta-go)(1)
Mặt khác: \(\frac{1}{\cos^2\partial}=\frac{1}{\frac{A^2}{H^2}}=\frac{H^2}{A^2}\)(2)
Vì (1) = (2) \(\Rightarrow1+\tan^2\partial=\frac{1}{cos^2\partial}\) (đpcm)
f) Ta có: \(1+\cot^2\partial=1+\left(\frac{A}{O}\right)^2=\frac{O^2}{O^2}+\frac{A^2}{O^2}=\frac{O^2+A^2}{O^2}=\frac{H^2}{O^2}\)(theo định lý Py-ta-go)(1)
Mặt khác: \(\frac{1}{\sin^2\partial}=\frac{1}{\frac{O^2}{H^2}}=\frac{H^2}{O^2}\)(2)
Vì (1) = (2) \(\Rightarrow1+\cot^2\partial=\frac{1}{\sin^2\partial}\)(đpcm)
