## Task A-3.2.

Determine all values that the expression

$$
\frac{1+\cos ^{2} x}{\sin ^{2} x}+\frac{1+\sin ^{2} x}{\cos ^{2} x}
$$

can take, where $x$ is a real number.
## Solution.

By reducing to a common denominator, we obtain

$$
\frac{1+\cos ^{2} x}{\sin ^{2} x}+\frac{1+\sin ^{2} x}{\cos ^{2} x}=\frac{\cos ^{2} x+\cos ^{4} x+\sin ^{2} x+\sin ^{4} x}{\sin ^{2} x \cos ^{2} x}
$$

Using the formula $\cos ^{2} x+\sin ^{2} x=1$, we find that this expression is equal to

$$
\frac{1+\cos ^{4} x+\sin ^{4} x}{\sin ^{2} x \cos ^{2} x}
$$

By completing the square and applying the formula $\sin ^{2} x+\cos ^{2} x=1$ again, we obtain the expression

$$
\frac{1+\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x}{\sin ^{2} x \cos ^{2} x}=\frac{2-2 \sin ^{2} x \cos ^{2} x}{\sin ^{2} x \cos ^{2} x}
$$

By factoring and using the formula $\sin 2 x=2 \sin x \cos x$, we obtain the expression

$$
\frac{2}{\sin ^{2} x \cos ^{2} x}-2=\frac{8}{\sin ^{2} 2 x}-2
$$

Since $\sin 2 x \in[-1,1]$, then $\sin ^{2} 2 x \in[0,1]$ and we conclude that the given expression can take all values in the interval $[6,+\infty\rangle$.