What is the value of $$\frac{{32{{\cos }^6}x - 48{{\cos }^4}x + 18{{\cos }^2}x - 1}}{{4\sin x\,\cos x\,\sin \left( {60 - x} \right)\cos \left( {60 - x} \right)\sin \left( {60 + x} \right)\cos \left( {60 + x} \right)}}?$$

Solution(By Examveda Team)

$$\eqalign{ & \frac{{32{{\cos }^6}x - 48{{\cos }^4}x + 18{{\cos }^2}x - 1}}{{4\sin x\, \times \cos x \times \,\sin \left( {60 - x} \right) \times \cos \left( {60 - x} \right) \times \sin \left( {60 + x} \right) \times \cos \left( {60 + x} \right)}} \cr & {\text{Adding }} + 4\,{\text{and }} - 4, \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 18{\cos ^2}x + 3 \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 24{\cos ^2}x - 6{\cos ^2}x + 3 \cr & \Rightarrow 32{\cos ^6}x - 4 - 48{\cos ^4}x + 24{\cos ^2}x - 3\cos 2x \cr & \Rightarrow 4{\left( {2{{\cos }^2}x - 1} \right)^3} - 3\cos 2x \cr & \left[ {\therefore \,\cos 2x = 2{{\cos }^2}x - 1} \right] \cr & \Rightarrow 4{\cos ^3}2x - 3\cos 2x \cr & \Rightarrow \cos 6x \cr} $$
\[\left[ \begin{align} & \because \,\cos 3x\to 4{{\cos }^{3}}x-3\cos x \\ & \text{and} \\ & \because \,\cos 6x\to 4{{\cos }^{3}}2x-3\cos 2x \\ \end{align} \right]\]
$$\eqalign{ & {\text{Now, }}4\sin x\,.\cos x.\,\sin \left( {60 - x} \right).\cos \left( {60 - x} \right).\sin \left( {60 + x} \right).\cos \left( {60 + x} \right) \cr & \left[ {\therefore \,\sin x.\sin \left( {60 - x} \right).\sin \left( {60 + x} \right) = \frac{1}{4}\sin 3x} \right] \cr & \left[ {\therefore \,\cos x.\cos \left( {60 - x} \right).\cos \left( {60 + x} \right) = \frac{1}{4}\cos 3x} \right] \cr & = \frac{{\cos 6x}}{{\frac{1}{4}\sin 3x.\cos 3x}} \cr & = \frac{2}{2} \times \frac{{4\cos 6x}}{{\sin 3x.\cos 3x}} \cr & = \frac{{8\cos 6x}}{{2\sin 3x.\cos 3x}} \cr & = \frac{{8\cos 6x}}{{\sin 6x}} \cr & = 8\cot 6x \cr} $$