Given complex number

$z=2-i\left(2\tan\frac{5\pi}{8}\right)=2-2i\tan\frac{5\pi}{8}$

Factor out $2$:

$z=2\left(1-i\tan\frac{5\pi}{8}\right)$

Simplify the tangent term

Using

$\tan\left(\pi-\alpha\right)=-\tan\alpha,$

we get

$\tan\frac{5\pi}{8}=\tan\left(\pi-\frac{3\pi}{8}\right)=-\tan\frac{3\pi}{8}$

Therefore,

Write in polar-friendly form

Use the identity

$1+i\tan\phi=\frac{\cos\phi+i\sin\phi}{\cos\phi}=\sec\phi\,(\cos\phi+i\sin\phi)$

with $\phi=\frac{3\pi}{8}$:

$z=2\sec\frac{3\pi}{8}\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}\right)$

Read modulus and argument

Comparing with polar form $z=r(\cos\theta+i\sin\theta)$,

$r=2\sec\frac{3\pi}{8},\qquad \theta=\frac{3\pi}{8}$

Match with the options

This corresponds to:

$\left(2\sec\frac{3\pi}{8},\frac{3\pi}{8}\right)$

which is Option B.

Verification with stored answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.