Cho \(\cos x = \frac{1}{5},\frac{\pi }{2} < x < \pi \).

a) \[\sin \frac{x}{2} = \frac{{\sqrt {10} }}{4}\].

b) \(\cos \frac{x}{2} = \frac{{\sqrt {15} }}{4}\).

c) \(\tan \frac{x}{2} = \frac{{\sqrt 6 }}{3}\).

d) \(\cot \frac{x}{2} = \frac{{\sqrt 6 }}{2}\).

\(\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \sin \frac{x}{2} > 0.\\\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 - \frac{1}{5}}}{2}} = \frac{{\sqrt {10} }}{5}\end{array}\).

\(\begin{array}{l}\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \cos \frac{x}{2} > 0.\\\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} = \sqrt {\frac{{1 + \frac{1}{5}}}{2}} = \frac{{\sqrt {15} }}{5}.\\\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{{\sqrt {10} }}{5}}}{{\frac{{\sqrt {15} }}{5}}} = \frac{{\sqrt 6 }}{3}.\\\cot \frac{x}{2} = \frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}} = \frac{{\frac{{\sqrt {15} }}{5}}}{{\frac{{\sqrt {10} }}{5}}} = \frac{{\sqrt 6 }}{2}.\end{array}\)