JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 11)
If
$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$
then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :
$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$
then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :
$$4 + 2\sqrt 3 $$
$$ - 2 + \sqrt 3 $$
$$ - 2 - \sqrt 3 $$
$$-\,\,4 - 2\sqrt 3 $$
Explanation
Given,
$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0
$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos2x) $$-$$ sinx(sin2x) = 0
$$ \Rightarrow $$$$\,\,\,$$ cos3x $$-$$ sin3 x = 0
$$ \Rightarrow $$$$\,\,\,$$ tan3x = 1
$$ \Rightarrow $$$$\,\,\,$$ tanx = 1
$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$
= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$
= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$
= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$
= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$
= $$-$$ 2 $$-$$ $$\sqrt 3 $$
$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0
$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos2x) $$-$$ sinx(sin2x) = 0
$$ \Rightarrow $$$$\,\,\,$$ cos3x $$-$$ sin3 x = 0
$$ \Rightarrow $$$$\,\,\,$$ tan3x = 1
$$ \Rightarrow $$$$\,\,\,$$ tanx = 1
$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$
= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$
= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$
= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$
= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$
= $$-$$ 2 $$-$$ $$\sqrt 3 $$
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