Đáp án:
$\begin{array}{l}
c){\tan ^2}a - {\sin ^2}a.{\tan ^2}a\\
= {\tan ^2}a.\left( {1 - {{\sin }^2}a} \right)\\
= {\tan ^2}a.{\cos ^2}a\\
\left( {do:{{\cos }^2}a + {{\sin }^2}a = 1} \right)\\
= \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}.{\cos ^2}a\\
= {\sin ^2}a
\end{array}$