If 0<x<π and cosx+sinx=12 then tanx =

sin x+cos x=121+sin2x=14sin2x=-342tan x1+tan2x=-348 tan x=-3-3tan2x3tan2x+8tan x+3=0tan x=-8±64-362(3) =-8±276=4+73

If 0<x<π and cosx+sinx=12 then tanx =

sin x+cos x=121+sin2x=14sin2x=-342tan x1+tan2x=-348 tan x=-3-3tan2x3tan2x+8tan x+3=0tan x=-8±64-362(3) =-8±276=4+73

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$I f 0 < x < π a n d \cos x + \sin x = \frac{1}{2} t h e n \tan x =$

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$- \frac{(4 + \sqrt{7})}{3}$

$\frac{1 + \sqrt{7}}{4}$

$\frac{1 - \sqrt{7}}{4}$

$\frac{4 - \sqrt{7}}{3}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \sin x + \cos x & = & \frac{1}{2} \\ 1 + \sin 2 x & = & \frac{1}{4} \\ \sin 2 x & = & \frac{- 3}{4} \\ \frac{2 \tan x}{1 + \tan^{2} x} & = & \frac{- 3}{4} \\ 8 \tan x & = & - 3 - 3 \tan^{2} x \\ 3 \tan^{2} x + 8 \tan x + 3 & = & 0 \\ \tan x & = & \frac{- 8 \pm \sqrt{64 - 36}}{2 (3)} \\ = & \frac{- 8 \pm 2 \sqrt{7}}{6} = (\frac{4 + \sqrt{7}}{3}) \end{array}$