4. It is known that $3 \operatorname{tg}^{2} \gamma+3 \operatorname{ctg}^{2} \gamma+\frac{2}{\sin ^{2} \gamma}+\frac{2}{\cos ^{2} \gamma}=19$. Find the value of the expression $\sin ^{4} \gamma-\sin ^{2} \gamma$.

Answer. $-\frac{1}{5}$.
Transform the given equality in the condition. On the domain of definition, it is equivalent to the following:

$$
\begin{aligned}
& \frac{3 \sin ^{2} \gamma}{\cos ^{2} \gamma}+\frac{3 \cos ^{2} \gamma}{\sin ^{2} \gamma}+\frac{2}{\sin ^{2} \gamma}+\frac{2}{\cos ^{2} \gamma}=19 \Leftrightarrow 3 \sin ^{4} \gamma+3 \cos ^{4} \gamma+2 \cos ^{2} \gamma+2 \sin ^{2} \gamma=19 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow \\
& \Leftrightarrow\left(3 \sin ^{4} \gamma+6 \sin ^{2} \gamma \cos ^{2} \gamma+3 \cos ^{4} \gamma\right)+2=25 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow 3\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)^{2}+2=25 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} \gamma \cos ^{2} \gamma=\frac{1}{5}
\end{aligned}
$$

Therefore, $\sin ^{4} Y-\sin ^{2} Y=\sin ^{2} \gamma\left(\sin ^{2} \gamma-1\right)=-\sin ^{2} \gamma \cos ^{2} Y=-\frac{1}{5}$.