a: Xét ΔABC vuông tại A có \(tanB=\dfrac{AC}{AB};cosB=\dfrac{AB}{BC}\)
\(1+tan^2B=1+\left(\dfrac{AC}{AB}\right)^2\)
\(=1+\dfrac{AC^2}{AB^2}=\dfrac{AB^2+AC^2}{AB^2}=\dfrac{BC^2}{AB^2}\)
\(=1:\dfrac{AB^2}{BC^2}=1:cos^2B=\dfrac{1}{cos^2B}\)
b: Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)
Xét ΔABH vuông tại H có \(sinB=\dfrac{AH}{AB};cosB=\dfrac{BH}{AB}\)
\(a\cdot sinB\cdot cosB\)
\(=BC\cdot\dfrac{AC}{BC}\cdot\dfrac{AH}{AB}=AH\)
\(a\cdot cos^2B=a\cdot cosB\cdot cosB=BC\cdot\dfrac{BH}{BA}\cdot\dfrac{AB}{BC}=BH\)
BH+CH=BC
=>\(CH+a\cdot cos^2B=a\)
=>\(CH=a\left(1-cos^2B\right)=a\cdot sin^2B\)
c: \(AH^2=\left(a\cdot sinB\cdot cosB\right)^2=a^2\cdot sin^2B\cdot cos^2B\)
\(=a\cdot sin^2B\cdot a\cdot cos^2B=BH\cdot CH\)
\(BC\cdot BH=a\cdot a\cdot cos^2B=a^2\cdot cos^2B=\left(a\cdot cosB\right)^2=AB^2\)
\(BC\cdot AH=a\cdot a\cdot sinB\cdot cosB=BC^2\cdot\dfrac{AB}{BC}\cdot\dfrac{AC}{BC}=AB\cdot AC\)
