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

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

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

a: \(\sin a+cosa=\sqrt2\)

=>\(\sqrt2\cdot\sin\left(a+\frac{\pi}{4}\right)=\sqrt2\)

=>\(\sin\left(a+\frac{\pi}{4}\right)=1\)

=>\(a+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
=>\(a=\frac{\pi}{4}+k2\pi\)

\(cosa=cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\sin a=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\tan a=\tan\left(\frac{\pi}{4}\right)=1\)

\(\cot a=\cot\left(\frac{\pi}{4}\right)=1\)

b: \(F=\sin^5a+cos^5a\)

\(=\sin^5\left(\frac{\pi}{4}\right)+cos^5\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt2}{2}\right)^5+\left(\frac{\sqrt2}{2}\right)^5\)

\(=\frac{4\sqrt2}{32}+\frac{4\sqrt2}{32}=\frac{8\sqrt2}{32}=\frac{\sqrt2}{4}\)

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

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

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

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

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

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

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

Đáp án cần chọn là: D

\(\sin^2\alpha+\cos^2\alpha=1\\ \Rightarrow\cos^2\alpha=1-0,6^2=0,64\\ \Rightarrow\cos\alpha=0,8=\dfrac{4}{5}\\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{0,6}{0,8}=\dfrac{3}{4}\\ \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{0,75}=\dfrac{4}{3}\)

Ta có: \(\tan a\cdot\cot a=1\)

=>\(\cot a=\frac{1}{2,4}=1:\frac{12}{5}=\frac{5}{12}\)

Ta có; \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+\left(\frac{12}{5}\right)^2=\frac{169}{25}=\left(\frac{13}{5}\right)^2\)

=>\(\frac{1}{cosa}=\frac{13}{5}\)

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

Ta có: \(\tan a=\frac{\sin a}{cosa}\)

=>\(\sin a=\frac{5}{13}\cdot2,4=\frac{12}{13}\)

Ta có : P = sin3 α + cos3 α = ( sinα + cosα) 3 - 3sin α.cosα(sinα + cosα)

Ta có (sin α + cos α) 2 = sin2α + cos2α +  2sinα.cosα = 1 + 24/25 = 49/25.

Vì sin α + cosα > 0  nên ta chọn sinα + cosα = 7/5.

Thay   vào P ta được 

`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`