Question

Tính

Answer 1

Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin75=cos15\Rightarrow sin^275=cos^215\)

Lại có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan\alpha=cot\left(90-\alpha\right)\Rightarrow tan23=cot67\)

Tương tự \(\Rightarrow cot37=tan53\)

Thế vào biểu thức 

\(\Rightarrow sin^215+sin^275+tan23-cot67-\dfrac{cot37}{tan53}=sin^215+cos^215-1\)

\(=1-1=0\)

Answer 2

\(sin^215^0+sin^275^0+tan23^0-cot67^0-\dfrac{cot37^0}{tan53^0}\)

\(=cos^275^0+\left(1-cos^275^0\right)+cot67^0-cot67^0-\dfrac{tan53^0}{tan53^0}\)

\(=1-1=0\)