3. (Easy/Average) Given that $\tan x+\cot x=8$, find the value of $\sqrt{\sec ^{2} x+\csc ^{2} x-\frac{1}{2} \sec x \csc x}$.
Answer: $2 \sqrt{15}$
Solution: Note that
$$
\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}=\frac{1}{\sin x \cos x} .
$$

This means that $\sin x \cos x=\frac{1}{8}$.
Now,
$$
\begin{aligned}
\sqrt{\sec ^{2} x+\csc ^{2} x-\frac{1}{2} \sec x \csc x} & =\sqrt{\frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x}-\frac{1}{2}\left(\frac{1}{\cos x}\right)\left(\frac{1}{\sin x}\right)} \\
& =\sqrt{\frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}-\left(\frac{1}{2}\right)(8)} \\
& ==\sqrt{64-4}=2 \sqrt{15} .
\end{aligned}
$$