Question

Cho \(\sin\alpha=\dfrac{3}{5}\) và 0 < α < \(\dfrac{\pi}{2}\). Hãy tính A=\(\dfrac{\cot\alpha+\tan\alpha}{cot\alpha-\tan\alpha}\)

Answer 1

Ta có: \(sin^2a+cos^2a=1\\ < =>cos^2a=1-\dfrac{9}{25}=\dfrac{16}{25}< =>cosa=\dfrac{4}{5}\left(0< a< \dfrac{\pi}{2}\right)\)

Ta có: \(A=\dfrac{cota+tana}{cota-tana}=\dfrac{\dfrac{cosa}{sina}+\dfrac{sina}{cosa}}{\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}=\dfrac{\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}+\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}}{\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}-\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}}=\dfrac{\dfrac{4}{3}+\dfrac{3}{4}}{\dfrac{4}{3}-\dfrac{3}{4}}=\dfrac{25}{7}\)