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Bài 1: 

\(\cos\alpha=\dfrac{4}{5}\)

\(\tan\alpha=\dfrac{3}{4}\)

\(\cot\alpha=\dfrac{4}{3}\)

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\)

\(\cot=\frac{1}{\tan}=\frac{1}{\frac{2\sqrt{5}}{5}}=\frac{\sqrt{5}}{2}\)

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 :)

ta có: tan B=\(\dfrac{8}{15}\)

=>tan B=\(\dfrac{8}{15}=\dfrac{AC}{AB}\)

mà AB=30 cm (gt)

=> AC= 8.30:15=16 cm

xét tam giác ABC vuông tại A (gt) 

=> AC²+AB²=BC²  ( Định lí pytago)

hay 16²+30²=BC²

=>BC=\(\sqrt{16^2+30^2}=34\)

ta có sin B=\(\dfrac{AC}{CB}=\dfrac{16}{34}=\dfrac{8}{17}\)

cos B= \(\dfrac{AB}{BC}=\dfrac{30}{34}=\dfrac{15}{17}\)

cotg B =\(\dfrac{30}{16}=\dfrac{15}{8}\)

\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)

\(=0+2\cdot1-2=0\)

\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)

\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)

\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)

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 \(\tan B=3\tan C\)

Chia cả tử và mẫu cho \(\sin a\)ta được

\(\frac{\sin a-\cos a}{\sin a+\cos a}=\frac{\frac{\sin a}{\sin a}-\frac{\cos a}{\sin a}}{\frac{\sin a}{\sin a}+\frac{\cos a}{\sin a}}=\frac{1-\cot a}{1+\cot a}=\frac{1-0,7456}{1+0,7456}=\frac{\frac{159}{625}}{\frac{1091}{625}}=\frac{159}{1091}\)

a: sin a=2/3

=>cos^2a=1-(2/3)^2=5/9

=>\(cosa=\dfrac{\sqrt{5}}{3}\)

\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)

\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

b: cos a=1/5

=>sin^2a=1-(1/5)^2=24/25

=>\(sina=\dfrac{2\sqrt{6}}{5}\)

\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

c: cot a=1/tana=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>1/cos^2a=1+4=5

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)

Ta có: \(\frac{cosa+sina}{cosa-sina}=\frac{\frac{cosa}{cosa}+\frac{sina}{cosa}}{\frac{cosa}{cosa}-\frac{sina}{cosa}}=\frac{1+tana}{1-tana}=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}=2\)