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Answer 1

$a, V ớ i 0^{o} < x < 90^{o} = > \sin x > 0; \cos x > 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \sin^{2} x = 1 - \cos^{2} x = 1 - {(\frac{3}{5})}^{2} = \frac{16}{25} = > \sin x = \frac{4}{5}, d o \sin x > 0 \tan x = \frac{\sin x}{\cos x} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} c o t x = \frac{1}{\tan x} = \frac{3}{4} v ậ y \sin x = \frac{4}{5}; \tan x = \frac{4}{3}; c o t x = \frac{3}{4} b, 90^{o} < x < 180^{o} = > \sin x > 0; \cos x < 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \sin^{2} x = 1 - \cos^{2} x = 1 - {(- \frac{5}{13})}^{2} = \frac{144}{169} = > \sin x = \sqrt{\frac{144}{169}} = \frac{12}{13}, d o \sin x > 0 = > \tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{- \frac{5}{13}} = - \frac{12}{5} c o t x = \frac{1}{\tan x} = - \frac{5}{12} V ậ y \sin x = \frac{12}{13}; \tan x = - \frac{12}{5}; c o t x = - \frac{5}{12}$

$c, V ớ i 90^{o} < x < 180^{o} = > \sin x > 0; \cos x < 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \cos^{2} x = 1 - \sin^{2} x = 1 - {(\frac{5}{13})}^{2} = \frac{144}{169} = > \cos x = - \sqrt{\frac{144}{169}} = - \frac{12}{13}, d o \cos x < 0 = > \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{- \frac{12}{13}} = - \frac{5}{12} c o t x = \frac{1}{\tan x} = - \frac{12}{5} v ậ y \cos x = - \frac{12}{13}; \tan x = - \frac{5}{12}; c o t x = - \frac{12}{5}$ ..

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Answer 2

$a, V ớ i 0^{o} < x < 90^{o} = > \sin x > 0; \cos x > 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \sin^{2} x = 1 - \cos^{2} x = 1 - {(\frac{3}{5})}^{2} = \frac{16}{25} = > \sin x = \frac{4}{5}, d o \sin x > 0 \tan x = \frac{\sin x}{\cos x} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} c o t x = \frac{1}{\tan x} = \frac{3}{4} v ậ y \sin x = \frac{4}{5}; \tan x = \frac{4}{3}; c o t x = \frac{3}{4} b, 90^{o} < x < 180^{o} = > \sin x > 0; \cos x < 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \sin^{2} x = 1 - \cos^{2} x = 1 - {(- \frac{5}{13})}^{2} = \frac{144}{169} = > \sin x = \sqrt{\frac{144}{169}} = \frac{12}{13}, d o \sin x > 0 = > \tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{- \frac{5}{13}} = - \frac{12}{5} c o t x = \frac{1}{\tan x} = - \frac{5}{12} V ậ y \sin x = \frac{12}{13}; \tan x = - \frac{12}{5}; c o t x = - \frac{5}{12}$

$c, V ớ i 90^{o} < x < 180^{o} = > \sin x > 0; \cos x < 0 T a c ó : \sin^{2} x + \cos^{2} x = 1 = > \cos^{2} x = 1 - \sin^{2} x = 1 - {(\frac{5}{13})}^{2} = \frac{144}{169} = > \cos x = - \sqrt{\frac{144}{169}} = - \frac{12}{13}, d o \cos x < 0 = > \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{- \frac{12}{13}} = - \frac{5}{12} c o t x = \frac{1}{\tan x} = - \frac{12}{5} v ậ y \cos x = - \frac{12}{13}; \tan x = - \frac{5}{12}; c o t x = - \frac{12}{5}$ ..

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