Cho tanx + cotx = 3. Tính A = sinx . cosx.

tanx + cotx = 3

\( \Leftrightarrow \) (tanx + cotx)² = 9

\( \Leftrightarrow \)tan²x + 2tanx.cotx + cot²x = 9

\( \Leftrightarrow \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} + 2\tan x.\cot x = 9\)

\( \Leftrightarrow \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^2}x.{{\sin }^2}x}} = 9 - 2\)

\( \Leftrightarrow \) sin⁴x + cos⁴x = 7cos²x.sin²x

\( \Leftrightarrow \)(sin²x + cos²x)² = 9sin²x.cos²x

\( \Leftrightarrow \)1 = 9sin²x.cos²x

\( \Leftrightarrow \)sin²x.cos²x = \(\frac{1}{9}\)

\( \Leftrightarrow \)\(\sin x.\cos x = \frac{1}{3}\)

Vậy \(A = \sin x\,\,.\,\,\cos x = \frac{1}{3}\).