Rút gọn: B= $\sqrt[]{2}$ – $\frac{1}{Sin(x+2013pi}$ .$\sqrt[]{1/(1+Cosx)+1/(1-Cosx)}$ 06/09/2021 Bởi aihong Rút gọn: B= $\sqrt[]{2}$ – $\frac{1}{Sin(x+2013pi}$ .$\sqrt[]{1/(1+Cosx)+1/(1-Cosx)}$
$\sqrt 2 – \dfrac{1}{{\sin \left( {x + 2013\pi } \right)}}\sqrt {\dfrac{1}{{1 + \cos x}} + \dfrac{1}{{1 – \cos x}}} = \sqrt 2 – \dfrac{1}{{\sin \left( {x + 2012\pi + \pi } \right)}}.\sqrt {\dfrac{{1 – \cos x + 1 + \cos x}}{{1 – {{\cos }^2}x}}} = \sqrt 2 – \dfrac{1}{{ – \sin x}}.\sqrt {\dfrac{2}{{{{\sin }^2}x}}} = \sqrt 2 + \dfrac{{\sqrt 2 }}{{\sin x|\sin x|}}$ Bình luận
$\sqrt 2 – \dfrac{1}{{\sin \left( {x + 2013\pi } \right)}}\sqrt {\dfrac{1}{{1 + \cos x}} + \dfrac{1}{{1 – \cos x}}} = \sqrt 2 – \dfrac{1}{{\sin \left( {x + 2012\pi + \pi } \right)}}.\sqrt {\dfrac{{1 – \cos x + 1 + \cos x}}{{1 – {{\cos }^2}x}}} = \sqrt 2 – \dfrac{1}{{ – \sin x}}.\sqrt {\dfrac{2}{{{{\sin }^2}x}}} = \sqrt 2 + \dfrac{{\sqrt 2 }}{{\sin x|\sin x|}}$