Đáp án:
\[\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} + \sqrt {\frac{{1 - \sin x}}{{1 + \sin x}}} = \frac{2}{{\cos x}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} + \sqrt {\frac{{1 - \sin x}}{{1 + \sin x}}} \\
= \frac{{\sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} }} + \frac{{\sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} }}\,\,\,\,\,\,\,\,\,\left( { - 1 \le \sin x \le 1} \right)\\
= \frac{{{{\sqrt {1 + \sin x} }^2} + {{\sqrt {1 - \sin x} }^2}}}{{\sqrt {1 - \sin x} .\sqrt {1 + \sin x} }}\\
= \frac{{\left( {1 + \sin x} \right) + \left( {1 - \sin x} \right)}}{{\sqrt {1 - {{\sin }^2}x} }}\\
= \frac{2}{{\sqrt {{{\cos }^2}x} }}\\
= \frac{2}{{\cos x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \in \left( { - \frac{\pi }{2};\frac{\pi }{2}} \right) \Rightarrow \cos x > 0} \right)
\end{array}\)
Vậy \(\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} + \sqrt {\frac{{1 - \sin x}}{{1 + \sin x}}} = \frac{2}{{\cos x}}\)
