\( \left. \begin{array} { l }{ \frac { \sin ^ { 2 } t } { \cos t } = \sec t - \cos t } \\ { \frac { \sin ^ { 2 } t } { \cos t } = \frac { \square } { \cos t } } \\ { = \square } \\ { = \square } \\ { = \sec t - \cos t } \end{array} \right. \)

\[ \begin{array}{l} \displaystyle \frac{\sin^2 t}{\cos t} = \sec t - \cos t \\[1mm] \displaystyle \frac{\sin^2 t}{\cos t} = \frac{1-\cos^2 t}{\cos t} \\[1mm] \displaystyle \frac{1-\cos^2 t}{\cos t} = \frac{1}{\cos t} - \cos t \\[1mm] \displaystyle \frac{1}{\cos t} - \cos t = \sec t - \cos t \\[1mm] \displaystyle \sec t - \cos t = \sec t - \cos t \end{array} \] To explain each step: 1. Replace \(\sin^2 t\) using the Pythagorean identity: \(\sin^2 t = 1-\cos^2 t\). 2. Write the resulting expression over the common denominator \(\cos t\). 3. Separate the fraction: \[ \frac{1-\cos^2 t}{\cos t} = \frac{1}{\cos t} - \frac{\cos^2 t}{\cos t} = \frac{1}{\cos t} - \cos t. \] 4. Recognize that \(\frac{1}{\cos t}\) is exactly \(\sec t\). Thus, the filled squares are: - First square: \(1-\cos^2t\) - Second square: \(\frac{1}{\cos t} - \cos t\) - Third square: \(\sec t-\cos t\)