Here, $$2n\alpha = \frac{\pi }{2}.$$   So, $$\cot \alpha \cdot \cot \left( {2n - 1} \right)\alpha = \cot \alpha \cdot \cot \left( {\frac{\pi }{2} - \alpha } \right) = \cot \alpha \cdot \tan \alpha = 1.$$

Value $$ = - \frac{{\cos {{126}^ \circ }}}{{\frac{1}{2}\sin {{126}^ \circ }}} = - 2\cot {126^ \circ } = 2\tan {36^ \circ } = \frac{{4 \cdot \sin {{18}^ \circ } \cdot \cos {{18}^ \circ }}}{{\cos {{36}^ \circ }}}$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4 \cdot \frac{{\sqrt 5 - 1}}{4} \cdot \sqrt {1 - {{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)}^2}} }}{{\frac{{\sqrt 5 + 1}}{4}}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\sqrt {10 + 2\sqrt 5 } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}\left( {10 + 2\sqrt 5 } \right)} }}{{\sqrt 5 + 1}}. \cr} $$

We take $$Q_3$$ first,
$$\eqalign{ & {Q_3} = \sin A\left( {\cos B + \cos C} \right) + \sin B\left( {\cos C + \cos A} \right) + \sin C\left( {\cos A + \cos B} \right) \cr & = \sin A\cos B + \sin A\cos C + \sin B\cos C + \sin B\cos A + \sin C\cos A + \sin C\cos B \cr & = \sin \left( {A + B} \right) + \sin \left( {B + C} \right) + \sin \left( {C + A} \right) = {Q_1} \cr & \Rightarrow {Q_3} = {Q_1} \cr} $$

The expression $$ = \left( {\sin {{78}^ \circ } - \sin {{42}^ \circ }} \right) - \left( {\sin {{66}^ \circ } - \sin {6^ \circ }} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cdot \cos {60^ \circ } \cdot \sin {18^ \circ } - 2 \cdot \cos {36^ \circ } \cdot \sin {30^ \circ }$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin {18^ \circ } - \cos {36^ \circ }.$$       Put values and simplify.

Given equation, $${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$
Since range of $$\sec \theta$$  is $$\left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right).$$
$$\eqalign{ & \therefore {\sec^2}\theta \geqslant 1 \cr & \Rightarrow \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}} \geqslant 1 \cr & \Rightarrow {\left( {x + y} \right)^2} - 4xy \leqslant 0 \cr & \Rightarrow {x^2} + {y^2} + 2xy - 4xy \leqslant 0 \cr} $$
$$ \Rightarrow {\left( {x - y} \right)^2} \leqslant 0$$    But $${\left(x-y\right)}^2\cancel{<}0$$   for any $$x,y \in R$$
$$\eqalign{ & \therefore {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x = y \cr} $$

We have,
$$\eqalign{ & 2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = 2\left[ {{{\left( {{{\sin }^2}\theta } \right)}^3} + {{\left( {{{\cos }^2}\theta } \right)}^3}} \right] - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = 2\left[ {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta \,{{\cos }^2}\theta } \right)} \right] - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = \left[ {2\,{{\sin }^4}\theta + 2\,{{\cos }^4}\theta - 2\,{{\sin }^2}\theta \,{{\cos }^2}\theta } \right] - 3\,{\sin ^4}\theta - 3\,{\cos ^4}\theta + 1 \cr & = - {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} + 1 \cr & = - {1^2} + 1 = - 1 + 1 = 0 \cr} $$

$$\eqalign{ & a = 2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2} - 2{\cos ^2}\frac{{A + B}}{2} + 1 \cr & a = 2\cos \frac{{A + B}}{2}\left\{ {\cos \frac{{A - B}}{2} - \cos \frac{{A + B}}{2}} \right\} + 1 \cr & a = 2\cos \frac{{A + B}}{2} \cdot 2\sin \frac{A}{2} \cdot \sin \frac{B}{2} + 1 = b + 1. \cr} $$

$$\eqalign{ & \sin \theta + {\text{cosec}} \, \theta = 2 \cr & \Rightarrow \,\,{\sin ^2}\theta - 2\sin \theta + 1 = 0 \cr & \Rightarrow \,\,\sin \theta = 1. \cr} $$

$$\eqalign{ & \sin \left\{ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \frac{\pi }{4}} \right\} \cr & = \sin \left\{ {\left( {\omega + {\omega ^2}} \right)\pi - \frac{\pi }{4}} \right\} \cr & = \sin \left( { - \pi - \frac{\pi }{4}} \right) \cr & = - \sin \left( {\pi + \frac{\pi }{4}} \right) \cr & = \sin \frac{\pi }{4} \cr & = \frac{1}{{\sqrt 2 }}. \cr} $$

$${2^x} + {2^{ - x}} \geqslant 2$$   (using AM $$ \geqslant $$ GM). But $$\sin \left( {{e^x}} \right) \leqslant 1.$$   So, there is no solution.