Cho cos a = 3/4, sina > 0, sin b = 3/5, cosb < 0.Khi đó:

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

a) Đúng. tan²a + 1 = \(\frac{1}{{{{\cos }^2}a}}\) nên \[\tan a = \sqrt {\frac{1}{{{{\cos }^2}a}} - 1} = \sqrt {\frac{{16}}{9} - 1} = \frac{{\sqrt 7 }}{3}\].

b) Sai. cot²b + 1 = \(\frac{1}{{{{\sin }^2}b}}\) nên \(\cot b = - \sqrt {\frac{1}{{{{\sin }^2}b}} - 1} = - \sqrt {\frac{{25}}{9} - 1} = \frac{{ - 4}}{3}\).

c) Sai. \(\left\{ {\begin{array}{*{20}{c}}{\cos a = \frac{3}{4}}\\{\sin a > 0}\end{array}} \right.\) \( \Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} = \frac{{\sqrt 7 }}{4}\)

và \(\left\{ {\begin{array}{*{20}{c}}{\sin b = \frac{3}{5}}\\{\cos b < 0}\end{array}} \right.\) \( \Rightarrow \cos b = - \sqrt {1 - {{\sin }^2}b} = \frac{{ - 4}}{5}\).

cos(a + b) = cosacosb – sinasinb = \(\frac{3}{4} \cdot \left( {\frac{{ - 4}}{5}} \right) - \frac{{\sqrt 7 }}{4} \cdot \frac{3}{5} = \frac{{ - 3}}{5} \cdot \left( {1 + \frac{{\sqrt 7 }}{4}} \right) \notin \left( {\frac{{ - 1}}{2};\frac{{ - 1}}{3}} \right)\).

d) Sai.\(\cos 2a + \cos 2b = 2{\cos ^2}a - 1 + 1 - 2{\sin ^2}b = 2 \cdot \frac{9}{{16}} - 2 \cdot \frac{9}{{25}} = \frac{{81}}{{200}} = 0,405 \notin \left( {\frac{1}{2};1} \right)\)