Take LHS:
\[sin\text{ }2x\text{ }/\text{ }\left( 1\text{ }+\text{ }cos\text{ }2x \right)\]
From formulae, \[cos\text{ }2x\text{ }=\text{ }1\text{ }\text{ }2\text{ }si{{n}^{2}}~x\]
\[=\text{ }2\text{ }co{{s}^{2}}~x\text{ }\text{ }1\]
\[Sin\text{ }2x\text{ }=\text{ }2\text{ }sin\text{ }x\text{ }cos\text{ }x\]
SO,
\[sin\text{ }2x\text{ }/\text{ }\left( 1\text{ }+\text{ }cos\text{ }2x \right)\text{ }=\text{ }[2\text{ }sin\text{ }x\text{ }cos\text{ }x\text{ }/\text{ }(1\text{ }+\text{ }(2co{{s}^{2}}x\text{ }\text{ }1))]\]
\[=\text{ }[2\text{ }sin\text{ }x\text{ }cos\text{ }x\text{ }/\text{ }(1\text{ }+\text{ }2co{{s}^{2}}~x\text{ }\text{ }1)]\]
\[=\text{ }[2\text{ }sin\text{ }x\text{ }cos\text{ }x\text{ }/\text{ }2\text{ }co{{s}^{2}}~x]\]
= sin x/cos x
= tan x
= RHS
Hence proved.