Let $$r$$ and $$\theta$$ respectively be the modulus and amplitude of the complex number $$z = 2 - i\left(2\tan\frac{5\pi}{8}\right)$$, then $$(r, \theta)$$ is equal to

We need to find the modulus $$r$$ and amplitude $$\theta$$ of the complex number $$z = 2 - i\left(2\tan\frac{5\pi}{8}\right)$$.

Note that $$\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$$. Using the identity $$\tan(\pi - x) = -\tan x$$, we get $$\tan\frac{5\pi}{8} = -\tan\frac{3\pi}{8}$$.

Substituting this into the expression for $$z$$ gives $$z = 2 - i\left(2 \times \left(-\tan\frac{3\pi}{8}\right)\right) = 2 + 2i\tan\frac{3\pi}{8}$$. Since $$\frac{3\pi}{8}$$ is in the first quadrant, $$\tan\frac{3\pi}{8} > 0$$, so $$z$$ has positive real and imaginary parts and lies in the first quadrant.

The modulus is $$r = |z| = \sqrt{4 + 4\tan^2\frac{3\pi}{8}} = 2\sqrt{1 + \tan^2\frac{3\pi}{8}}$$. Using the identity $$1 + \tan^2 x = \sec^2 x$$ gives $$r = 2\sqrt{\sec^2\frac{3\pi}{8}} = 2\sec\frac{3\pi}{8}$$, which is positive since $$\sec\frac{3\pi}{8} > 0$$.

For the amplitude, $$\tan\theta = \frac{\mathrm{Im}(z)}{\mathrm{Re}(z)} = \frac{2\tan\frac{3\pi}{8}}{2} = \tan\frac{3\pi}{8}$$. Because $$z$$ lies in the first quadrant and $$\frac{3\pi}{8} \in \bigl(0,\frac{\pi}{2}\bigr)$$, it follows that $$\theta = \frac{3\pi}{8}$$.

Therefore, $$(r, \theta) = \left(2\sec\frac{3\pi}{8},\; \frac{3\pi}{8}\right)$$.

Answer: Option 1 — $$\left(2\sec\frac{3\pi}{8}, \frac{3\pi}{8}\right)$$