sin20<sin70

cos25 > cos65*15'

tan73*20' >tan45

cotg2 >cotg73*40'

tan25>sin25

cotg32 >cos32

bài 1 : ta có : \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(0,6\right)^2=\dfrac{16}{25}\)

\(\Rightarrow cosa=\pm\dfrac{4}{5}\)

\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\pm\dfrac{3}{4}\) \(\Rightarrow cotx=\dfrac{1}{tanx}=\pm\dfrac{4}{3}\)

bài 2)

ý 1 : a) ta có : \(\dfrac{1}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=tan^2a+1\left(đpcm\right)\)

b) ta có : \(\dfrac{1}{sin^2a}=\dfrac{sin^2a+cos^2a}{sin^2a}=1+cot^2a\left(đpcm\right)\)

c) \(cos^4a-sin^4a=\left(sin^2a+cos^2a\right)\left(cos^2a-sin^2a\right)\)

\(=cos^2a-sin^2a=2cos^2a-cos^2a-sin^2a=2cos^2a-1\left(đpcm\right)\)

ý 2 :

ta có : \(tana=2\Rightarrow cota=\dfrac{1}{2}\)

ta có : \(tan^2a+1=\dfrac{1}{cos^2a}\Leftrightarrow cos^2a=\dfrac{1}{tan^2a+1}=\dfrac{1}{5}\)

\(\Rightarrow cosa=\pm\dfrac{1}{\sqrt{5}}\Rightarrow sin^2a=1-cos^2a=\dfrac{4}{5}\) \(\Rightarrow sina=\pm\dfrac{2}{\sqrt{5}}\)

vậy ............................................................................

bài 3 bạn tự luyện tập như bài 2 cho quen nha :)

sin 39 ° 13 '  ≈ 0,6323     cos 52 ° 18 '  ≈ 0,6115

tg 13 ° 20 '  ≈ 0,2370     cotg 10 ° 17 '  ≈ 0,5118

sin 45 °  ≈ 0,7071     cos 45 °  ≈ 0,7071

Chọn đáp án A

1. Ta có \(\tan a=3\Rightarrow\frac{\sin a}{\cos a}=3\Rightarrow\sin a=3\cos a\)

Vậy \(\frac{\cos a+\sin a}{\cos a-\sin a}=\frac{\cos a+3\cos a}{\cos a-3\cos a}=\frac{4\cos a}{-2\cos a}=-2\)

2.Ta có \(\sin^2a+\cos^2a=1\Rightarrow\cos^2a=1-\sin^2a=1-\frac{4}{9}=\frac{5}{9}\)

\(\Rightarrow\orbr{\begin{cases}\cos a=\frac{\sqrt{5}}{3}\\\cos a=\frac{-\sqrt{5}}{3}\end{cases}}\)

Với \(\cos a=\frac{\sqrt{5}}{3}\Rightarrow\tan a=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\Rightarrow\cot a=\frac{1}{\tan a}=\frac{\sqrt{5}}{2}\)

Với \(\cos a=\frac{-\sqrt{5}}{2}\Rightarrow\tan a=\frac{-2\sqrt{5}}{5}\Rightarrow\cot a=-\frac{\sqrt{5}}{2}\)

3. 

Theo hệ thức  lượng trong tam giác vuông ta có \(AB^2=BH.BC\Leftrightarrow10^2=5.BC\Rightarrow BC=20\left(cm\right)\)

Theo định lí Pitago thì \(AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)\)

Ta có \(\tan B=\frac{AC}{AB}=\frac{10\sqrt{3}}{10}=\sqrt{3};\tan C=\frac{AB}{AC}=\frac{1}{\sqrt{3}}\)

Vậy \(\tan B=3\tan C\)