JEE MAIN - Mathematics (2006 - No. 15)
If $$0 < x < \pi $$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is :
$${{\left( {1 - \sqrt 7 } \right)} \over 4}$$
$${{\left( {4 - \sqrt 7 } \right)} \over 3}$$
$$ - {{\left( {4 + \sqrt 7 } \right)} \over 3}$$
$${{\left( {1 + \sqrt 7 } \right)} \over 4}$$
Explanation
$$\cos x + \sin x = {1 \over 2}$$
$$ \Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$
$$ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$
$$ \Rightarrow 1 + \sin 2x = {1 \over 4}$$
$$ \Rightarrow \sin 2x = - {3 \over 4},$$ 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 $$\tan x < 0\,$$
$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$
$$ \Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$
$$ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$
$$ \Rightarrow 1 + \sin 2x = {1 \over 4}$$
$$ \Rightarrow \sin 2x = - {3 \over 4},$$ 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 $$\tan x < 0\,$$
$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$
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