sin a=12/13

cos^2a=1-(12/13)^2=25/169

=>cosa=5/13

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

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

sin b=căn 3/2

cos^2b=1-(căn 3/2)^2=1/4

=>cos b=1/2

tan b=căn 3/2:1/2=căn 3

cot b=1/căn 3

\(\cos\alpha=0.8\)

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

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

\(sina=0,6\Rightarrow cosa=\sqrt{1-sin^2a}=\sqrt{1-0,6^2}=0,8\)

\(tana=\dfrac{sina}{cosa}=\dfrac{0,6}{0,8}=\dfrac{3}{4}\)

\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)

a) sin a=0,8

Ta có: \(\sin^2a+\cos^2a=1\)

\(\Rightarrow\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\orbr{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

\(\sin a=0,8\)

\(\sin^2a=1-\sin^2a=1\)

\(\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\hept{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

Code : Breacker

Câu 1:

\(\sin\widehat{B}=\dfrac{12}{13}\)

\(\cos\widehat{B}=\dfrac{5}{13}\)

\(\tan\widehat{B}=\dfrac{12}{5}\)

\(\cot\widehat{B}=\dfrac{5}{12}\)

a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

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

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

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

Bài 1:

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC^2=6^2-4^2=36-16=20\)

=>\(AC=2\sqrt5\) (cm)

Xét ΔABC vuông tại A có

sin B=cos C=\(\frac{AC}{BC}=\frac{2\sqrt5}{6}=\frac{\sqrt5}{3}\)

cos B=sin C=\(\frac{AB}{BC}=\frac46=\frac23\)

tan B=cot C=\(\frac{AC}{AB}=\frac{2\sqrt5}{4}=\frac{\sqrt5}{2}\)

cot B=tan C=\(\frac{AB}{AC}=\frac{4}{2\sqrt5}=\frac{4\sqrt5}{10}=\frac{2\sqrt5}{5}\)

BÀi 2:

a: \(A=cos^2x+cos^2x\cdot\cot^2x\)

\(=cos^2x\left(1+\cot^2x\right)\)

\(=\frac{cos^2x}{\sin^2x}=\cot^2x\)

b: \(\sin^2x+\sin^2x\cdot\tan^2x\)

\(=\sin^2x\left(1+\tan^2x\right)\)

\(=\sin^2x:cos^2x=\tan^2x\)