Cho $\sin a=\frac{5}{13}$; $\tan b=\frac{3}{4}$, với $a,b$ là các góc nhọn.

a) Do $a$ là góc nhọn $\Rightarrow \cos a=\sqrt{1-{\sin }^{2}a}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13}\Rightarrow$ a) Sai.
b) $\sin 2a=2\sin a\cos a=2.\frac{5}{13}.\frac{12}{13}=\frac{120}{169}\Rightarrow$ b) Sai.
c) $\tan a=\frac{5}{12}\Rightarrow$ c) Đúng.
d) $\tan a=\frac{5}{12},\tan b=\frac{3}{4}\Rightarrow \tan (a+b)=\frac{56}{33}\Rightarrow$ d) Đúng.
a) Do $a$ là góc nhọn $\Rightarrow \cos a=\sqrt{1-{\sin }^{2}a}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13}\Rightarrow$ a) Sai.
b) $\sin 2a=2\sin a\cos a=2.\frac{5}{13}.\frac{12}{13}=\frac{120}{169}\Rightarrow$ b) Sai.
c) $\tan a=\frac{5}{12}\Rightarrow$ c) Đúng.
d) $\tan a=\frac{5}{12},\tan b=\frac{3}{4}\Rightarrow \tan (a+b)=\frac{56}{33}\Rightarrow$ d) Đúng.