The reason B is wrong is this.

\(\displaystyle \alpha < \beta,\ \dfrac{ \pi }{4} \le \alpha < \pi, \text { and } \dfrac{ \pi }{4} < \beta \le \pi \implies\)

\(\displaystyle -\ 1 < cos( \alpha ) \le \dfrac{\sqrt{2}}{2} \text { and } 0 \le sin ( \beta ) < \dfrac{\sqrt{2}}{2}.\)

So, **for example**,

\(\displaystyle \alpha = \dfrac{\pi}{4} \text { and } \beta = \pi \implies \alpha < \beta,\ cos( \alpha) = \dfrac{\sqrt{2}}{2}, \text { and } sin ( \beta ) = 0 \implies\)

\(\displaystyle \alpha < \beta \text { and } cos ( \alpha ) >> sin ( \beta ).\)

It is a tricky problem because you may think that alpha and beta must be in the same quadrant, but nothing in the problem requires that. If alpha is in the first quadrant and beta is in the second, both the cosine of alpha and the sine of beta **will **be positive, and the cosine of alpha **may** exceed the sine of beta. It is of course true that if both were in the first quadrant or both were in the second quadrant, your reasoning would be correct, but the problem does not specify that.