Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \dfrac{1}{{\sin x - \sqrt {{{\cot }^2}x - {{\cos }^2}x} }}\\
= \dfrac{1}{{\sin x - \sqrt {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - {{\cos }^2}x} }}\\
= \dfrac{1}{{\sin x - \sqrt {{{\cos }^2}x.\left( {\dfrac{1}{{{{\sin }^2}x}} - 1} \right)} }}\\
= \dfrac{1}{{\sin x - \sqrt {{{\cos }^2}x.\dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}}} }}\\
= \dfrac{1}{{\sin x - \sqrt {{{\cos }^2}x,\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} }}\\
= \dfrac{1}{{\sin x - \left| {\dfrac{{{{\cos }^2}x}}{{\sin x}}} \right|}}\\
= \dfrac{1}{{\sin x + \dfrac{{{{\cos }^2}x}}{{\sin x}}}}\,\,\,\,\,\left( {\pi < x < 2\pi \Rightarrow \sin x < 0} \right)\\
= \dfrac{1}{{\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x}}}} = \dfrac{1}{{\dfrac{1}{{\sin x}}}} = \sin x\\
b,\\
0 < x < \dfrac{\pi }{2} \Rightarrow 0 < \sin x,\cos x < 1\\
P = \sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} - \sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} \\
= \dfrac{{\sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} }} - \dfrac{{\sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} }}\\
= \dfrac{{{{\sqrt {1 + \sin x} }^2} - {{\sqrt {1 - \sin x} }^2}}}{{\sqrt {1 - \sin x} .\sqrt {1 + \sin x} }}\\
= \dfrac{{\left( {1 + \sin x} \right) - \left( {1 - \sin x} \right)}}{{\sqrt {1 - {{\sin }^2}x} }}\\
= \dfrac{{2\sin x}}{{\sqrt {{{\cos }^2}x} }} = \dfrac{{2\sin x}}{{\cos x}} = 2\tan x
\end{array}\)
