For $0 < \theta < \dfrac{\pi }{2}$, the solution of $\sum\limits_{m = 1}^6 {{\text{cosec}}\left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right){\text{cosec}}\left( {\theta + \dfrac{{m\pi }}{4}} \right)} = 4\sqrt 2 $ is/are?
$\left( a \right)\dfrac{\pi }{4}$
$\left( b \right)\dfrac{\pi }{6}$
$\left( c \right)\dfrac{\pi }{{12}}$
$\left( d \right)\dfrac{{5\pi }}{{12}}$

Hint: In this particular question use the concept that cosec x = (1/sin x), then expand the summation and use the property that $\dfrac{1}{{\sin x\sin \left( {x + a} \right)}} = \dfrac{1}{{\sin a}}\left[ {\cot x - \cot \left( {x + a} \right)} \right]$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given expression
$\sum\limits_{m = 1}^6 {{\text{cosec}}\left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right){\text{cosec}}\left( {\theta + \dfrac{{m\pi }}{4}} \right)} = 4\sqrt 2 $, for $0 < \theta < \dfrac{\pi }{2}$
Now we have to find out the values of $\theta $
As we know that cosec x = (1/sin x) so use this property in the above expression we have,
$ \Rightarrow \sum\limits_{m = 1}^6 {\dfrac{1}{{\sin \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{m\pi }}{4}} \right)}}} = 4\sqrt 2 $
Now expand the summation we have,
$ \Rightarrow \dfrac{1}{{\sin \theta \sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{1}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{{2\pi }}{4}} \right)}} + ........... + \dfrac{1}{{\sin \left( {\theta + \dfrac{{5\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{6\pi }}{4}} \right)}} = 4\sqrt 2 $..... (1)
Now, simplify the above expression using the property which is given as,
$ \Rightarrow \dfrac{1}{{\sin x\sin \left( {x + a} \right)}} = \dfrac{1}{{\sin a}}\left[ {\dfrac{{\sin \left( {x + a - x} \right)}}{{\sin x\sin \left( {x + a} \right)}}} \right]$
$ \Rightarrow \dfrac{1}{{\sin x\sin \left( {x + a} \right)}} = \dfrac{1}{{\sin a}}\left[ {\dfrac{{\sin \left( {x + a} \right)\cos x - \cos \left( {x + a} \right)\sin x}}{{\sin x\sin \left( {x + a} \right)}}} \right]$, $\left[ {\because \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B} \right]$
Now simplify we have,
$ \Rightarrow \dfrac{1}{{\sin x\sin \left( {x + a} \right)}} = \dfrac{1}{{\sin a}}\left[ {\dfrac{{\cos x}}{{\sin x}} - \dfrac{{\cos \left( {x + a} \right)}}{{\sin \left( {x + a} \right)}}} \right]$
$ \Rightarrow \dfrac{1}{{\sin x\sin \left( {x + a} \right)}} = \dfrac{1}{{\sin a}}\left[ {\cot x - \cot \left( {x + a} \right)} \right]$, where cot x = (cos x/sin x)
So use this property in equation (1), where a = $\dfrac{\pi }{4}$ so we have,
$ \Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\cot \theta - \cot \left( {\theta + \dfrac{\pi }{4}} \right) + \cot \left( {\theta + \dfrac{\pi }{4}} \right) - \cot \left( {\theta + \dfrac{{2\pi }}{4}} \right) + ...... + \cot \left( {\theta + \dfrac{{5\pi }}{4}} \right) - \cot \left( {\theta + \dfrac{{6\pi }}{4}} \right)} \right] = 4\sqrt 2 $
So as we see that except first and last term all the terms are cancel out so we have,

$ \Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\cot \theta - \cot \left( {\theta + \dfrac{{6\pi }}{4}} \right)} \right] = 4\sqrt 2 $
$ \Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\cot \theta - \cot \left( {\theta + \dfrac{{3\pi }}{2}} \right)} \right] = 4\sqrt 2 $
$ \Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\cot \theta - \cot \left( {\pi + \left( {\theta + \dfrac{\pi }{2}} \right)} \right)} \right] = 4\sqrt 2 $
Now as we know that $\cot \left( {\pi + x} \right) = \cot x$, as cot is positive in the third quadrant.
$ \Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\cot \theta - \cot \left( {\dfrac{\pi }{2} + \theta } \right)} \right] = 4\sqrt 2 $
Now as we know that $\cot \left( {\dfrac{\pi }{2} + x} \right) = - \tan x,\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ so use this property we have,
$ \Rightarrow \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}\left[ {\cot \theta + \tan \theta } \right] = 4\sqrt 2 $
$ \Rightarrow \left[ {\cot \theta + \tan \theta } \right] = 4$
$ \Rightarrow \left[ {\dfrac{1}{{\tan \theta }} + \tan \theta } \right] = 4$
$ \Rightarrow {\tan ^2}\theta - 4\tan \theta + 1 = 0$
So this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow \tan \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Where, a = 1, b = -4, c = 1.
So we have,
$ \Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {{4^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}} = \dfrac{{4 \pm \sqrt {12} }}{2} = 2 \pm \sqrt 3 $
$ \Rightarrow \tan \theta = \left( {2 + \sqrt 3 } \right),\left( {2 - \sqrt 3 } \right)$
So when, $\tan \theta = \left( {2 + \sqrt 3 } \right)$
$ \Rightarrow \theta = \dfrac{{5\pi }}{{12}} = {75^o}$
And when, $\tan \theta = \left( {2 - \sqrt 3 } \right)$
$ \Rightarrow \theta = \dfrac{\pi }{{12}} = {15^o}$
So this is the required answer.
Hence options (c) and (d) are the correct answer.

Note:Whenever we face such types of questions the key concept we have to remember is that always recall basic trigonometric identities such as $\left[ {\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B} \right]$, $\cot \left( {\pi + x} \right) = \cot x$, $\cot \left( {\dfrac{\pi }{2} + x} \right) = - \tan x,\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$, and always recall the formula to solve the complex quadratic equation which is stated above.