Question

CMR nếu α+β=90 độ thì sin α =cosβ, tan α= cot β

 

Answer 1

\(sin\alpha=cos\beta=\dfrac{AB}{BC}\)

\(tan\alpha=cot\beta=\dfrac{AB}{AC}\)

Answer 2

\(\alpha+\beta=90^o\)

\(\Rightarrow\beta=90^o-\alpha\)

Theo đề bài :

\(sin\alpha=cos\beta\)

\(\Rightarrow sin\alpha=cos\left(90^o-\alpha\right)\)

mà \(\alpha;90^o-\alpha\) là 2 góc phụ nhau

\(\Rightarrow cos\left(90^o-\alpha\right)=sin\alpha\left(dpcm\right)\)

Tương tự \(tan\alpha=cot\beta=cot\left(90^o-\alpha\right)\)