Cho sinx = 3/5 với pi/2 < x < π khi đó tan(x + pi/4) bằng?

Đáp án đúng là: D

\[{\cos ^2}x = 1 - {\sin ^2}x = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}\].

Do đó \(\cos x = \frac{4}{5}\), \(\cos x = \frac{{ - 4}}{5}\). Vì \(\frac{\pi }{2}\) < x < π nên \(\cos x = \frac{{ - 4}}{5}\)

\[\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\frac{3}{5}}}{{\frac{{ - 4}}{5}}} = \frac{{ - 3}}{4}\].

\(\tan \left( {x + \frac{\pi }{4}} \right) = \frac{{\tan x + \tan \frac{\pi }{4}}}{{1 - \tan x\tan \frac{\pi }{4}}} = \frac{{\frac{{ - 3}}{4} + 1}}{{1 - \left( {\frac{{ - 3}}{4}} \right) \cdot 1}} = \frac{{\frac{1}{4}}}{{\frac{7}{4}}} = \frac{1}{7}\).