Question

chứng minh

Answer 1

Đặt \(\widehat{B}=\alpha\)

a: \(sin^2\alpha+cos^2\alpha\)

\(=sin^2B+cos^2B\)

\(=\left(\dfrac{AC}{BC}\right)^2+\left(\dfrac{AB}{BC}\right)^2=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)

b: \(tan\alpha=tanB=\dfrac{AC}{AB}\)

\(=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{sina}{cosa}\)

c: \(cota=cotB\)

\(=\dfrac{AB}{AC}\)

\(=\dfrac{AB}{BC}:\dfrac{AC}{BC}=\dfrac{cosa}{sina}\)

d: \(tana\cdot cota=\dfrac{cosa}{sina}\cdot\dfrac{sina}{cosa}=1\)

Answer 2

\(a)sin^2\alpha+cos^2\alpha=1\\ \Leftrightarrow\left(\dfrac{đối}{huyền}\right)^2+\left(\dfrac{kề}{huyền}\right)^2=1\\ \Leftrightarrow\dfrac{đối^2}{huyền^2}+\dfrac{kề^2}{huyền^2}=1\\ \Leftrightarrow\dfrac{đối^2+kề^2}{huyền^2}=1\\ \Leftrightarrow\dfrac{huyền^2}{huyền^2}=1\left(pytago\right)\\ \Leftrightarrow1=1\left(đpcm\right)\)

\(b)tan\alpha=\dfrac{sin\alpha}{cos\alpha}\\ \Leftrightarrow tan\alpha=\dfrac{đối}{huyền}:\dfrac{kề}{huyền}\\ \Leftrightarrow tan\alpha=\dfrac{đối\cdot huyên}{huyền\cdot kề}\\ \Leftrightarrow tan\alpha=\dfrac{đối}{kề}\\ \Leftrightarrow tan\alpha=tan\alpha\left(đpcm\right)\)

\(c)cotg\alpha=\dfrac{cos\alpha}{sin\alpha}\\ \Leftrightarrow cotg\alpha=\dfrac{kề}{huyền}:\dfrac{đối}{huyền}\\ \Leftrightarrow cotg\alpha=\dfrac{kề\cdot huyền}{huyền\cdotđối}\\ \Leftrightarrow cotg\alpha=\dfrac{kề}{đối}\\ \Leftrightarrow cotg\alpha=cotg\alpha\left(đpcm\right)\)

\(d)tan\alpha\cdot cotg\alpha=1\\ \Leftrightarrow\dfrac{đối}{kề}\cdot\dfrac{kề}{đối}=1\\ \Leftrightarrow\dfrac{đối\cdot kề}{kề\cdotđối}=1\\ \Leftrightarrow1=1\left(đpcm\right)\)