(i) \[tan\text{ }{{36}^{o}}~+\text{ }tan\text{ }{{9}^{o}}~+\text{ }tan\text{ }{{36}^{o}}~tan\text{ }{{9}^{o}}~=~1\]
We know \[36{}^\circ \text{ }+\text{ }9{}^\circ \text{ }=\text{ }45{}^\circ \]
So we have,
\[tan\text{ }\left( 36{}^\circ \text{ }+\text{ }9{}^\circ \right)\text{ }=\text{ }tan\text{ }45{}^\circ \]
Since, \[tan\text{}\left(A\text{}+\text{}B\right)\text{}=\text{}\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}A\text{}tan\text{}B\right)\]
Therefore,
\[(tan\text{ }{{36}^{o}}+\text{ }tan\text{ }{{9}^{o}})\text{ }/\text{ }(1\text{}-\text{}tan\text{}{{36}^{o}}tan\text{}{{9}^{o}})\text{}=\text{}1\]
\[tan\text{ }36{}^\circ \text{ }+\text{ }tan\text{ }9{}^\circ \text{ }=\text{ }1\text{}-\text{}tan\text{}36{}^\circ\text{}tan\text{}9{}^\circ\]
Rearanging we get,
\[tan\text{ }36{}^\circ \text{ }+\text{ }tan\text{ }9{}^\circ \text{ }+\text{ }tan\text{ }36{}^\circ \text{ }tan\text{ }9{}^\circ \text{ }=\text{ }1\]
Hence proved.
(ii) \[tan\text{}13x\text{}-\text{}tan\text{}9x\text{}-\text{}tan\text{}4x\text{ }=\text{}tan\text{}13x\text{}tan\text{}9x\text{}tan\text{}4x\]
Let us consider LHS:
\[tan\text{ }13x\text{ }\text{ }tan\text{ }9x\text{ }\text{ }tan\text{ }4x\]
\[tan\text{ }13x\text{ }=\text{ }tan\text{ }\left( 9x\text{ }+\text{ }4x \right)\]
Since, \[tan\text{}\left(A\text{}+\text{}B\right)\text{}=\text{}\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}A\text{}tan\text{}B\right)\]
Therefore,
\[tan\text{}13x\text{}=\text{}\left(tan\text{}9x\text{}+\text{}tan\text{}4x\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}9x\text{ }tan\text{}4x \right)\]
By cross-multiplying we get,
\[tan\text{}13x\text{}\left(1\text{}-\text{}tan\text{}9x\text{}tan\text{}4x\right)\text{}=\text{}tan\text{}9x\text{}+\text{}tan\text{}4x\]
\[tan\text{ }13x\text{ }-\text{ }tan\text{ }13x\text{ }tan\text{ }9x\text{ }tan\text{ }4x\text{ }=\text{ }tan\text{ }9x\text{ }+\text{ }tan\text{ }4x\]
Upon rearranging we get,
\[tan\text{ }13x\text{ }-\text{ }tan\text{ }9x\text{ }-\text{ }tan\text{ }4x\]
\[=\text{ }tan\text{ }13x\text{ }tan\text{ }9x\text{ }tan\text{ }4x\]
\[=\text{ }RHS\]
\[\therefore LHS\text{ }=\text{ }RHS\]
Hence proved.