Question

\(\sqrt{\dfrac{1+sinx}{1-sinx}}+\sqrt{\dfrac{1-sinx}{1+sinx}}=?\) (sao cho gọn nhất)

Answer 1

\(=\dfrac{1+sinx+1-sinx}{\sqrt{\left(1-sinx\right)\left(1+sinx\right)}}=\dfrac{2}{\sqrt{1-sin^2x}}=\dfrac{2}{\sqrt{cos^2x}}=\dfrac{2}{\left|cosx\right|}\)