\(sin^2\alpha+cos^2\alpha=1\Rightarrow cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(0,6\right)^2}=\frac{4}{5}\)

\(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{0,6}{\frac{4}{5}}=\frac{3}{4}\)

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

Điều kiện: a>45 độ

\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)

\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)

\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

\(\cot\alpha=\dfrac{1}{2}\)

\(\sin\alpha=\dfrac{kề}{\sqrt{5}kề}=\dfrac{\sqrt{5}}{5}\)

\(\cos\alpha=\sqrt{1-\dfrac{5}{25}}=\dfrac{2\sqrt{5}}{5}\)

a: Xét ΔABC vuông tại A có tan B=\(\frac{AC}{AB}\) ; cos B=\(\frac{BA}{BC}\)

\(1+\tan^2B=1+\left(\frac{AC}{AB}\right)^2=\frac{AB^2+AC^2}{AB^2}=\frac{BC^2}{AB^2}\)

\(\frac{1}{cos^2B}=1:\left(\frac{BA}{BC}\right)^2=1:\frac{BA^2}{BC^2}=\frac{BC^2}{BA^2}\)

Do đó: \(1+\tan^2B=\frac{1}{cos^2B}\)

b:

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC;AB^2=BH\cdot BC;AC^2=CH\cdot CB\)

Xét ΔABC vuông tại A có sin B=\(\frac{AC}{BC};cosB=\frac{AB}{BC}\)

\(a\cdot\sin B\cdot cosB\)

\(=BC\cdot\frac{AB}{BC}\cdot\frac{AC}{BC}=\frac{AH\cdot BC}{BC}=AH\)

Xét ΔABC vuông tại A có cos B=\(\frac{BA}{BC}\)

=>\(cos^2B=\frac{BA^2}{BC^2}=\frac{BH\cdot BC}{BC^2}=\frac{BH}{BC}\)

=>\(BH=BC\cdot cos^2B=a\cdot cos^2B\)

TA có: BH+CH=BC

=>\(CH=a-a\cdot cos^2B=a\left(1-cos^2B\right)=a\cdot\sin^2B\)

\(\dfrac{\pi}{2}< a< \pi\Rightarrow sina>0\)

\(\Rightarrow sina=\sqrt{1-cos^2a}=\dfrac{\sqrt{5}}{3}\)

\(K=2sina.cosa+2cos^2a-1=-\dfrac{1}{9}-\dfrac{4}{9}\sqrt{5}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{1}{4}\Rightarrow a-b=-3\)

\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)