a) $tg{28}^{\circ }=\frac{\sin {28}^{\circ }}{\cos {28}^{\circ }}=\sin {28}^{\circ }.\frac{1}{\cos {28}^{\circ }}$ (1)

Vì 0 < cos28° < 1 nên $\frac{1}{\cos {28}^{\circ }}>1\Rightarrow \sin {28}^{\circ }.\frac{1}{\cos {28}^{\circ }}>\sin {28}^{\circ }$ (2)

Từ (1) và (2) suy ra: tg28° > sin28°

b) Ta có: $\cot g{42}^{\circ }=\frac{\cos {42}^{\circ }}{\sin {42}^{\circ }}=c{os42}^{\circ }.\frac{1}{\sin {42}^{\circ }}$ (1)

Vì 0 < sin42° < 1 nên $\frac{1}{\sin {42}^{\circ }}>1\Rightarrow \cos$

a: \(\sin25^0< \sin70^0\)

b: \(\cos40^0>\cos75^0\)

c: \(\sin38^0=\cos52^0< \cos27^0\)

d: \(\sin50^0=\cos40^0>\cos50^0\)

a: \(\tan50^028'< \tan63^0\)

b: \(\cot14^0>\cot35^012'\)

c: \(\tan27^0=\cot63^0< \cot27^0\)

d: \(\tan65^0=\cot25^0>\cot65^0\)

a, \(\sin25^0\)< \(\sin70^0\)

b, \(\cos40^0\)> \(\cos75^0\)

c, \(\sin35^0\)= \(\cos55^0\)

\(\cos55^0\)< \(\cos35^0\)

\(\Rightarrow\)\(\sin35^0\)< \(\cos35^0\)

#mã mã#

a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)

                  \(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)

                    \(\cos\alpha\)= 3 \(\sin\alpha\)

ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)

#mã mã#

a/ Có \(\sin B=\frac{AC}{BC};\sin C=\frac{AB}{BC};\cos B=\frac{AB}{BC};\cos C=\frac{AC}{BC}\)

\(\Rightarrow\frac{\sin B-\sin C}{\cos B-\cos C}=\frac{AC-AB}{AB-AC}\)

Nếu AC<AB=> AC-AB<0 =>...<0

Nếu AC>AB=>AB-AC<0=>...<0

b/ làm tg tự câu a

c/ \(\cot B=\frac{AB}{AC};\cot C=\frac{AC}{AB}\)

\(\Rightarrow\cot B+\cot C=\frac{AB^2+AC^2}{AB.AC}\)

Quy đồng lên có: \(AB^2+AC^2>2AB.AC\) (luôn đúng vs AB\(\ne\) AC)

Vậy đẳng thức đc CM