sin x < 0 .

Do \(0 < x < \frac{\pi }{2}\) nên \(\sin x > 0\).

Ta có \({\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x = 1 - \frac{1}{{16}} = \frac{{15}}{{16}}\)

\( \Rightarrow \sin x = \frac{{\sqrt {15} }}{4};\tan x = \frac{{\sin x}}{{\cos x}} = \sqrt {15} ;\cot x = \frac{{\cos x}}{{\sin x}} = \frac{1}{{\sqrt {15} }}.\)

Đáp án:       a) Sai,                    b) Sai,                    c) Đúng,          d) Sai.