Đặt tên hình như hình sau:
Ta có:
\(\begin{array}{l}\sin \alpha = \dfrac{a}{c} \Rightarrow {\sin ^2}a = \dfrac{{{a^2}}}{{{c^2}}}\\\cos \alpha = \dfrac{b}{c} \Rightarrow {\cos ^2}\alpha = \dfrac{{{b^2}}}{{{c^2}}}\\{\sin ^2}a + {\cos ^2}\alpha = \dfrac{{{a^2}}}{{{c^2}}} + \dfrac{{{b^2}}}{{{c^2}}} = \dfrac{{{c^2}}}{{{c^2}}} = 1\end{array}\)
Vậy đáp án là (C)
Bài 2.10
(A) \(tg\alpha = \sin\alpha + \cos\alpha\) ;
(B) \(tg\alpha = \sin\alpha - \cos\alpha\) ;
(C) \(tg\alpha = \sin\alpha .\cos\alpha\) ;
(D) \(tg \alpha\) = \(\dfrac{{\sin \alpha }}{{\cos \alpha }}.\)
\(\begin{array}{l}\sin \alpha = \dfrac{a}{c};{\rm{cos}}\alpha {\rm{ = }}\dfrac{b}{c}\\ \Rightarrow \dfrac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = \dfrac{a}{c}:\dfrac{b}{c} = \dfrac{a}{b} = tg\alpha \end{array}\)
Vậy đáp án (D)
Bài 2.11
(A) \(cotg\alpha = 1 + tg\alpha\);
(B) \(cotg\alpha = 1 − tg\alpha\);
(C) \(cotg\alpha = 1.tg\alpha\) ;
(D) \(cotg\alpha = \dfrac{1}{{tg\alpha }}.\)
\(\begin{array}{l}tg\alpha = \dfrac{a}{b};cotg\,\alpha = \dfrac{b}{a}\\tg\alpha .cotg\,\alpha = 1 \Rightarrow cotg\,\alpha = \dfrac{1}{{tg\alpha }}.\end{array}\)
Vậy đáp án là (D).