Let us consider LHS:

\[(tan\text{ }{{69}^{o}}~+\text{ }tan\text{ }{{66}^{o}})\text{ }/\text{ }(1\text{ }\text{ }tan\text{ }{{69}^{o}}~tan\text{ }{{66}^{o}})\]

Since, \[tan\text{ }\left( A\text{ }+\text{ }B \right)\text{}=\text{}\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}/\text{}\left(1\text{ }\text{ }tan\text{ }A\text{ }tan\text{ }B \right)\]

Here, \[A\text{ }=\text{ }{{69}^{o}}~and\text{ }B\text{ }=\text{ }{{66}^{o}}\]

Therefore,

\[(tan\text{ }{{69}^{o}}~+\text{ }tan\text{ }{{66}^{o}})\text{ }/\text{ }(1\text{ }\text{}tan\text{}{{69}^{o}}-tan\text{}{{66}^{o}})\text{}=\text{}tan\text{}{{\left(69\text{}+\text{}66\right)}^{o}}\]

\[=\text{ }tan\text{ }{{135}^{o}}\]

\[=\text{ }\text{ }tan\text{ }{{45}^{o}}\]

\[=\text{ }\text{ }1\]

\[=\text{ }RHS\]

\[\therefore LHS\text{ }=\text{ }RHS\]

Hence proved.