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

\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)

\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)

`sin^2 α+cos^2α=1`

`<=> (2/3)^2+cos^2α=1`

`=> cosα= \sqrt5/3`

`=> tan α=(sinα)/(cosα) = (2\sqrt5)/5`

`=> cota = 1/(tanα)=sqrt5/2`

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

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

dạ e cảm ơn nh ạ!!!!

Ta có: \(tana+cota=3\Rightarrow\dfrac{sina}{cosa}+\dfrac{cosa}{sina}=3\)

\(\Rightarrow\dfrac{sin^2a+cos^2a}{sina\cdot cosa}=3\Rightarrow sina\cdot cosa=\dfrac{1}{3}\)

Ta có: \(\left(tana+cota\right)^2=9\)\(\Rightarrow tan^2a+cot^2a=9-2tana\cdot cota=9-2=7\)

Cos^2(a) = 1/(1+tan^2(a)) = 4/13

--> cosa = 2sqrt(13)/13

Sin^2(a)=1-4/13=9/13

--> sina = 3sqrt(13)/13

Cota = 2/3 --> tana = 3/2

b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)

hay \(\cos\alpha=\dfrac{4}{5}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)

\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)

\(=\dfrac{141}{25}\)

c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)

D

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