If [0

Question

If $$0 < x < \pi $$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is :

Accepted Answer

$$\cos x + \sin x = {1 \over 2}$$

$$ \Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$

$$ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$

$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$

$$ \Rightarrow 1 + \sin 2x = {1 \over 4}$$

$$ \Rightarrow \sin 2x =  - {3 \over 4},$$ so $x$ is obtuse and

$${{2\tan x} \over {1 + {{\tan }^2}x}} =  - {3 \over 4}$$

$$ \Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$$

$$\therefore$$ $$\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$$

$$ = {{ - 4 \pm \sqrt 7 } \over 3}$$

as $$\tan x < 0\,$$

$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$

Solutions · 1