tan x=2
=>\(cotx=\dfrac{1}{tanx}=\dfrac{1}{2}\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+4=5\)
=>\(cos^2x=\dfrac{1}{5}\)
=>\(cosx=\dfrac{1}{\sqrt{5}}\) hoặc \(cosx=-\dfrac{1}{\sqrt{5}}\)
\(tanx=2\)
=>\(\dfrac{sinx}{cosx}=2\)
=>\(sinx=2\cdot cosx\)
TH1: \(cosx=\dfrac{1}{\sqrt{5}}\)
=>\(sinx=\dfrac{2}{\sqrt{5}}\)
\(H=\dfrac{sinx+cotx}{5sinx-5cosx}\)
\(=\dfrac{\dfrac{2}{\sqrt{5}}+\dfrac{1}{2}}{5\left(\dfrac{2}{\sqrt{5}}-\dfrac{1}{\sqrt{5}}\right)}=\dfrac{4+\sqrt{5}}{2\sqrt{5}}:\sqrt{5}=\dfrac{4+\sqrt{5}}{10}\)
TH2: \(cosx=-\dfrac{1}{\sqrt{5}}\)
=>\(sinx=2\cdot cosx=-\dfrac{2}{\sqrt{5}}\)
\(H=\dfrac{sinx+cotx}{5\left(sinx-cosx\right)}\)
\(=\dfrac{\dfrac{-2}{\sqrt{5}}+\dfrac{1}{2}}{5\left(-\dfrac{2}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}\right)}=\dfrac{-4+\sqrt{5}}{2\sqrt{5}}:\left(-\sqrt{5}\right)\)
\(=\dfrac{4-\sqrt{5}}{10}\)
