Ta có:

\(\tan \frac{\pi }{14}=\frac{1}{a}\Rightarrow \sin \frac{\pi }{7}\)\(=\frac{\frac{2}{a}}{1+\frac{1}{{{a}^{2}}}}\)\(=\frac{2a}{{{a}^{2}}+1}\)\(=\sin \frac{6\pi }{7};\cos \frac{\pi }{7}\)\(=\frac{1-\frac{1}{{{a}^{2}}}}{1+\frac{1}{{{a}^{2}}}}\)\(=\frac{{{a}^{2}}-1}{{{a}^{2}}+1}\)

\(\Rightarrow \sin \frac{2\pi }{7}\)\(=2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\)\(=\frac{4a\left( {{a}^{2}}-1 \right)}{{{\left( {{a}^{2}}+1 \right)}^{2}}}\)

\(\sin \frac{4\pi }{7}\)\(=\sin \frac{3\pi }{7}\)\(=3\sin \frac{\pi }{7}-4{{\sin }^{3}}\frac{\pi }{7}\)

\(\Rightarrow \sin \frac{4\pi }{7}+\sin \frac{6\pi }{7}\)\(=4\sin \frac{\pi }{7}-4{{\sin }^{3}}\frac{\pi }{7}=4\sin \frac{\pi }{7}\left( 1-{{\sin }^{2}}\frac{\pi }{7} \right)\)

\(=4\sin \frac{\pi }{7}.{{\cos }^{2}}\frac{\pi }{7}\)\(=2\sin \frac{2\pi }{7}\cos \frac{\pi }{7}\)

Khi đó:

\(K=\sin \frac{2\pi }{7}\left( 2\cos \frac{\pi }{7}+1 \right)\)\(=\frac{4a\left( {{a}^{2}}-1 \right)}{{{\left( {{a}^{2}}+1 \right)}^{2}}}.\frac{2{{a}^{2}}-2+{{a}^{2}}+1}{{{a}^{2}}+1}\)\(=\frac{4a\left( {{a}^{2}}-1 \right)\left( 3{{a}^{2}}-1 \right)}{{{\left( {{a}^{2}}+1 \right)}^{3}}}\)

Đáp án C.