1.\(sin^2\alpha+cos^2\alpha=\left(\dfrac{AC}{BC}\right)^2+\left(\dfrac{AB}{BC}\right)^2\)
=\(\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2\left(pytago\right)}{BC^2}=1\)
2.ta có \(tan\alpha=\dfrac{AC}{AB}\)
\(\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{AC}{BC}}{\dfrac{AB}{BC}}=\dfrac{AC}{AB}\)
\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
3.ta có:\(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2\)
=\(\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\)=\(\dfrac{1}{cos^2\alpha}\)
4.ta có :\(cot\alpha=\dfrac{AB}{AC}\)
\(\dfrac{cos\alpha}{sin\alpha}=\dfrac{\dfrac{AB}{BC}}{\dfrac{AC}{BC}}=\dfrac{AB}{AC}\)
\(\Rightarrow cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)
\(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}\)=\(\dfrac{1}{sin^2a}\)
