a) Vì 20 °   <   70 °   n ê n   sin   20 °   <   sin 70 °  (góc tăng, sin tăng)

b) Vì 25 °   <   63 ° 15 '   n ê n   cos 25 °   >   cos   63 ° 15 ' (góc tăng, cos giảm)

c) Vì 73 ° 20 '   >   45 °   n ê n   t g 73 ° 20 '   >   t g 45 °  (góc tăng, tg tăng)

d) Vì 2 °   <   37 ° 40 '   n ê n   c o t g   2 °   >   c o t g   37 ° 40 '  (góc tăng, cotg giảm )

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 :)

1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

\(1+cot^2a=\dfrac{1}{sin^2a}\)

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

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

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)

cos = \(\frac{\sqrt{3}}{2}\)

tan = \(\frac{1}{\sqrt{3}}\)

cot = \(\sqrt{3}\)

a/ \(\sin\alpha=\frac{C_đ}{C_h}\)

\(\cos\alpha=\frac{C_k}{C_h}\)

\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{C_đ}{C_h}}{\frac{C_k}{C_h}}=\frac{C_đ}{C_k}=\tan\alpha\)

b/ \(\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{C_k}{C_h}}{\frac{C_đ}{C_h}}=\frac{C_k}{C_đ}=\cot\alpha\)

c/ \(\tan\alpha.\cot\alpha=\frac{C_đ}{C_k}.\frac{C_k}{C_đ}=1\)

d/ \(\sin^2\alpha=\frac{C_đ^2}{C_h^2}\)

\(\cos^2\alpha=\frac{C_k^2}{C_h^2}\)

\(\Rightarrow\sin^2\alpha+\cos^2\alpha=\frac{C_đ^2+C_k^2}{C_h^2}=\frac{C_h^2}{C_h^2}=1\)

P/s: hok trc lp 9 hay sao mà lm bài bài này?