1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

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

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

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

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

\(\frac{1-tana}{1+tana}=\frac{1-\frac{sina}{cosa}}{1+\frac{sina}{cosa}}=\frac{\frac{1}{cosa}\left(cosa-sina\right)}{\frac{1}{cosa}\left(cosa+sina\right)}=\frac{cosa-sina}{cosa+sina}\)

xin lỗi mk ko thể giúp bn đc mk mới hc lp 7 thôi!

a) Mình nghĩ là cos a = cot a . sin a chứ :))

CM nà :

Ta có : cot a =  \(\frac{AB}{AC}\)(1)

\(\frac{cosa}{sina}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{AC}\left(2\right)\)

Từ (1) và (2)  \(\Rightarrow\)cot a =  \(\frac{cosa}{sina}\)

\(\Leftrightarrow\)cos a = cot a . sin a

b) Ta có : tan a =  \(\frac{AC}{AB}\)

Lại có : cot a =  \(\frac{AB}{AC}\)

\(\Rightarrow\)cos a . tan a =  \(\frac{AC.AB}{AB.AC}\)= 1 

Vậy ...

bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)

\(=cos^220+cos^270+cos^240+cos^250\)

\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)

\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)

bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)

\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)

b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)

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

Bài 1 :

\(C=cos^2a\left(cos^2a+sin^2a\right)+sin^2a=cos^2a+sin^2a=1\)

D = \(\left(sin^2a+cos^2a\right)+\left(cos\left(90-a\right)-sina\right)+1+\left(tan^2\left(90-a\right)-\frac{1}{sin^2a}\right)\)

  \(=1+\left(sina-sina\right)+1+\left(cot^2a-1-cos^2a\right)=1+1-1=1\)

tớ cần câu trả lời

a: tan α=3/4

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

=>\(\frac{1}{cos^2a}=1+\left(\frac34\right)^2=1+\frac{9}{16}=\frac{25}{16}\)

=>\(cos^2a=\frac{16}{25}\)

=>\(cosa=\sqrt{\frac{16}{25}}=\frac45\)

\(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac{16}{25}=\frac{9}{25}\)

=>sin a=\(\sqrt{\frac{9}{25}}=\frac35\)

b: \(1+\cot^2a=\frac{1}{\sin^2a}\)

=>\(\frac{1}{\sin^2a}=1+\left(\frac{5}{12}\right)^2=1+\frac{25}{144}=\frac{169}{144}\)

=>\(\sin^2a=\frac{144}{169}\)

=>\(\sin a=\sqrt{\frac{144}{169}}=\frac{12}{13}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-\frac{144}{169}=\frac{25}{169}=\left(\frac{5}{13}\right)^2\)

=>\(cosa=\frac{5}{13}\)

\(sin\alpha^2+cos\alpha^2=1\Rightarrow sin\alpha^2=1-cos\alpha^2=1-\dfrac{1}{25}=\dfrac{24}{25}\Rightarrow sin\alpha=\dfrac{2\sqrt{6}}{5}\)

\(\Rightarrow cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{24}\)

\(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\sin^2\alpha=1-\dfrac{1}{25}=\dfrac{24}{25}\)

hay \(\sin\alpha=\dfrac{2\sqrt{6}}{5}\)

\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

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