a) +) TH1: \(0 < \alpha < {90^0}\).
Xét tam giác \(ABC\) vuông tại A, đặt \(\widehat B = \alpha \) ta có:
\(\sin \alpha = \dfrac{{AC}}{{BC}},\cos \alpha = \dfrac{{AB}}{{BC}}\)
\( \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = \dfrac{{A{C^2}}}{{B{C^2}}} + \dfrac{{A{B^2}}}{{B{C^2}}}\) \( = \dfrac{{A{C^2} + A{B^2}}}{{B{C^2}}} = \dfrac{{B{C^2}}}{{B{C^2}}} = 1\).
+) TH2: \(\alpha = 0\) thì \(\sin \alpha = 0,\cos \alpha = 1\) nên \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
+) TH3: \(\alpha = {90^0}\) thì \(\sin \alpha = 1,\cos \alpha = 0\) nên \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
+) TH4: \({90^0} < \alpha < {180^0}\) thì \(0 < {180^0} - \alpha < {90^0}\) nên:
\({\sin ^2}\left( {{{180}^0} - \alpha } \right) + {\cos ^2}\left( {{{180}^0} - \alpha } \right) = 1 \Leftrightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
+) TH5: \(\alpha = {180^0}\) thì \(\sin \alpha = 0,\cos \alpha = - 1\) nên \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
Vậy với mọi \(\alpha \in \left[ {0;{{180}^0}} \right]\) ta đều có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
b) Ta có: \(\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\) nên \(1 + {\tan ^2}\alpha = 1 + \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \dfrac{1}{{{{\cos }^2}\alpha }}\)
