Take LHS:

\[1\text{ }+\text{ }co{{s}^{2}}~2x\]

From formula, \[cos2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\]

\[co{{s}^{2}}~x\text{ }+\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\]

So,

\[1\text{ }+\text{ }co{{s}^{2}}~2x\text{ }=\text{ }{{(co{{s}^{2}}~x\text{ }+\text{ }si{{n}^{2}}~x)}^{~2}}~+\text{ }{{(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)}^{~2}}\]

\[=\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\text{ }+\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]

\[=\text{ }co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\]

\[=\text{ }2\text{ }co{{s}^{4}}~x\text{ }+\text{ }2\text{ }si{{n}^{4}}~x\]

\[=\text{ }2\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x)\]

= RHS

Hence proved.