\[{\text{Докажите тождество:}}\]
$%\frac{{{{\left( {\sin t + \cos t} \right)}^2} - 1}}{{{\mathop{\rm tg}\nolimits} t - \sin t\cos t}} = 2{{\mathop{\rm ctg}\nolimits} ^2}t$%
\[\begin{array}{l}\frac{{{{\left( {\sin t + \cos t} \right)}^2} - 1}}{{{\mathop{\rm tg}\nolimits} t - \sin t\cos t}} = \\\left[ \begin{array}{l}{\left( {\sin t + \cos t} \right)^2} - 1 = \\{\sin ^2}t + {\cos ^2}t - 1 + 2\sin t\cos t = \\1 - 1 + 2\sin t\cos t = 2\sin t\cos t\end{array} \right] = \\\frac{{2\sin t\cos t}}{{{\mathop{\rm tg}\nolimits} t - \sin t\cos t}} = \\\left[ \begin{array}{l}{\mathop{\rm tg}\nolimits} t - \sin t\cos t = \frac{{\sin t}}{{\cos t}} - \sin t\cos t = \\\sin t\left( {\frac{1}{{\cos t}} - \cos t} \right) = \sin t \cdot \frac{{1 - {{\cos }^2}t}}{{\cos t}} = \\\sin t \cdot \frac{{{{\sin }^2}t}}{{\cos t}} = \frac{{{{\sin }^3}t}}{{\cos t}}\end{array} \right] = \\\frac{{2\sin t\cos t}}{{\frac{{{{\sin }^3}t}}{{\cos t}}}} = \frac{{2\sin t{{\cos }^2}t}}{{{{\sin }^3}t}} = \frac{{2{{\cos }^2}t}}{{{{\sin }^2}t}} = 2{{\mathop{\rm ctg}\nolimits} ^2}t\end{array}\]