QJEEM17Ma70a5: JEE Main 2017 Mathematics - Class 12 JEEM17Ma70a5 Solution: If S = { x ∈ [ 0 , 2 π ] :

Question

Verified Solution • 2025–26 Session moderate

If

S = {x \in [0, 2 π] : | \begin{matrix} 0 & \cos x & - \sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{matrix} | = 0},

then

\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)}

is equal to :

Accepted Answer

Given,

| \begin{matrix} 0 & \cos x & - \sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{matrix} |

= 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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