Answer
$(1-\cos 4x)(2 + \tan^2 x+\cot^2 x)=8$
Work Step by Step
Here, we have $(1-\cos 4x)(2 + \tan^2 x+\cot^2 x)=(1-\cos 4x)(2 + \tan^2 x+\dfrac{1}{\tan^2 x})$
or, $(1-\cos 4x)(1+\tan^2 x)^2(\dfrac{1}{\tan^2 x})=(1-\cos 4x)(1+\dfrac{1-\cos 2x}{1+\cos 2x})^2(\dfrac{1}{\tan^2 x})$
or, $(1-\cos 4x)(1+\dfrac{1-\cos 2x}{1+\cos 2x})^2(\dfrac{1}{\tan^2 x})=(1-\cos 4x)(\dfrac{2}{1+\cos 2x})^2(\dfrac{1}{\tan^2 x})$
or, $(1-\cos 4x)(\dfrac{2}{1+\cos 2x})^2(\dfrac{1}{\tan^2 x})=(1-\cos 4x)\dfrac{4}{(1+\cos 2x)^2}\cdot \dfrac{1}{\tan^2 x}$
or,$(1-\cos 4x)\dfrac{4}{(1+\cos 2x)^2}\cdot \dfrac{1}{\tan^2 x}=2(\dfrac{1-\cos 4x}{2})\dfrac{4}{(1+\cos 2x)^2}\cdot \dfrac{1}{\tan^2 x}$
or, $\dfrac{8\sin^2 x}{(1+\cos 2x)^2}\cdot \dfrac{1}{\tan^2 x}=\dfrac{8(1-\cos 2x)}{(1+\cos 2x)}\cdot \dfrac{1}{\tan^2 x}$
or,$\dfrac{8(1-\cos 2x)}{(1+\cos 2x)}\cdot \dfrac{1}{\tan^2 x}=\dfrac{8\tan^2 x}{\tan^2 x}$
Thus, $(1-\cos 4x)(2 + \tan^2 x+\cot^2 x)=8$
Hence, the result has been proved.
