Question

If 0 < x < π and cos x + sin x = $\frac{1}{2}$, then tan x is
(a) $\frac{(1-\sqrt{7})}{4}$
(b) $\frac{(4-\sqrt{7})}{3}$
(c) $-\frac{(4+\sqrt{7})}{3}$
(d) $\frac{(1+\sqrt{7})}{4}$

Answer 1

$cosx + sinx = \frac{1}{2} \\ \\divide \: with \: cosx \\ \\ 1 + tanx = \frac{1}{2} secx \\ \\ 2 + 2tanx = secx \\ \\ sobs \\ \\ 4 {tan}^{2} x + 8tanx + 4 = {sec}^{2} x \\ \\4 {tan}^{2} x + 8tanx + 4 = 1 + {tan}^{2} x \\ \\ 3{tan}^{2} x + 8tanx + 3 = 0 \\ \\ tanx = \frac{ - 8 + | \sqrt{64 - 4(3)(3)}| }{6} \\ \\ tanx = \frac{ - 4 - \sqrt{7} }{3} \\ \\ tanx = \frac{ - 4 + \sqrt{7} }{3}$

hope it helps