Prove the following identities: \[\mathbf{sin}\text{ }\mathbf{2x}\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]
Prove the following identities: \[\mathbf{sin}\text{ }\mathbf{2x}\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

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.