Question
If $$0 < x < \pi \,\,{\text{and}}$$ $$\cos x + \sin x = \frac{1}{2},$$ then tan$$x$$ is
A.
$$\frac{{\left( {1 - \sqrt 7 } \right)}}{4}$$
B.
$$\frac{{\left( {4 - \sqrt 7 } \right)}}{3}$$
C.
$$ - \frac{{\left( {4 + \sqrt 7 } \right)}}{3}$$
D.
$$\frac{{\left( {1 + \sqrt 7 } \right)}}{4}$$
Answer :
$$ - \frac{{\left( {4 + \sqrt 7 } \right)}}{3}$$
Solution :
$$\eqalign{
& \cos x + \sin x = \frac{1}{2} \cr
& \Rightarrow \,\,1 + \sin 2x = \frac{1}{4} \cr} $$
$$ \Rightarrow \,\,\sin 2x = - \frac{3}{4},$$ so $$x$$ is obtuse and $$\frac{{2\tan x}}{{1 + {{\tan }^2}x}} = - \frac{3}{4}$$
$$\eqalign{
& \Rightarrow \,\,3{\tan ^2}x + 8\tan x + 3 = 0 \cr
& \therefore \,\,\tan x = \frac{{ - 8 \pm \sqrt {64 - 36} }}{6} \cr
& = - \frac{{ - 4 \pm \sqrt 7 }}{3} \cr
& {\text{as }}\tan x < 0 \cr
& \therefore \,\,\tan x = \frac{{ - 4 - \sqrt 7 }}{3} \cr} $$