4. It is known that $4 \operatorname{tg}^{2} Y+4 \operatorname{ctg}^{2} Y-\frac{1}{\sin ^{2} \gamma}-\frac{1}{\cos ^{2} \gamma}=17$. Find the value of the expression $\cos ^{2} Y-\cos ^{4} \gamma$.

Answer. $\frac{3}{25}$.
Transform the given equality in the condition. On the domain of definition, it is equivalent to the following:

$$
\begin{aligned}
\frac{4 \sin ^{2} \gamma}{\cos ^{2} \gamma}+\frac{4 \cos ^{2} \gamma}{\sin ^{2} \gamma}-\frac{1}{\sin ^{2} \gamma}-\frac{1}{\cos ^{2} \gamma}= & 17 \Leftrightarrow 4 \sin ^{4} \gamma+4 \cos ^{4} \gamma-\cos ^{2} \gamma-\sin ^{2} \gamma=17 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow \\
\Leftrightarrow\left(4 \sin ^{4} \gamma+8 \sin ^{2} \gamma \cos ^{2} \gamma+4 \cos ^{4} \gamma\right)-1 & =25 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow 4\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)^{2}-1=25 \sin ^{2} \gamma \cos ^{2} \gamma \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} \gamma \cos ^{2} \gamma=\frac{3}{25} .
\end{aligned}
$$

Therefore, $\cos ^{2} \gamma-\cos ^{4} \gamma=\cos ^{2} \gamma\left(1-\cos ^{2} \gamma\right)=\sin ^{2} \gamma \cos ^{2} \gamma=\frac{3}{25}$.