\[\left( \mathbf{i} \right)tan\text{ }8x\text{ }-\text{ }tan\text{ }6x\text{ }-\text{ }tan\text{ }2x\text{ }=\text{}tan\text{}8x\text{}tan\text{ }6x\text{ }tan\text{ }2x\]
Let us consider LHS:
\[tan\text{ }8x\text{ }\text{ }tan\text{ }6x\text{ }\text{ }tan\text{ }2x\]
\[tan\text{ }8x\text{ }=\text{ }tan\left( 6x\text{ }+\text{ }2x \right)\]
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,
\[tan\text{ }8x\text{ }=\text{ }\left( tan\text{ }6x\text{ }+\text{}tan\text{}2x\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}6x\text{}tan\text{}2x\right)\]
By cross-multiplying we get,
\[tan\text{ }8x\text{ }\left( 1\text{ }-\text{ }tan\text{ }6x\text{ }tan\text{ }2x \right)\text{ }=\text{ }tan\text{ }6x\text{ }+\text{ }tan\text{ }2x\]
\[tan\text{ }8x\text{}-\text{}tan\text{}8x\text{}tan\text{}6x\text{}tan2x\text{}=\text{ }tan\text{ }6x\text{ }+\text{ }tan\text{ }2x\]
Rearranging we get,
\[tan\text{}8x\text{ }-\text{ }tan\text{ }6x\text{ }-\text{ }tan\text{ }2x\text{ }=\text{}tan\text{}8x\text{}tan\text{}6x\text{}tan\text{ }2x\]
\[=\text{ }RHS\]
\[\therefore LHS\text{ }=\text{ }RHS\]
Hence proved.
(ii) \[tan\text{ }\pi /12\text{ }+\text{ }tan\text{ }\pi /6\text{ }+\text{ }tan\text{ }\pi /12\text{ }tan\text{ }\pi /6\text{ }=\text{ }1\]
We know,
\[\pi /12\text{ }=\text{ }15{}^\circ \text{ }and\text{ }\pi /6\text{ }=\text{ }30{}^\circ \]
Therefore, we have \[15{}^\circ \text{ }+\text{ }30{}^\circ \text{ }=\text{ }45{}^\circ \]
\[Tan\text{ }\left( 15{}^\circ \text{ }+\text{ }30{}^\circ \right)\text{ }=\text{ }tan\text{ }45{}^\circ \]
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,
\[(tan\text{}{{15}^{o}}+\text{}tan\text{}{{30}^{o}})\text{}/\text{}(1\text{}-\text{}tan\text{}{{15}^{o}}tan\text{}{{30}^{o}})\text{}=\text{ }1\]
\[tan\text{}15{}^\circ \text{}+\text{}tan\text{}30{}^\circ\text{}=\text{}1\text{}-\text{}tan\text{}15{}^\circ\text{}tan\text{}30{}^\circ\]
Upon rearranging we get,
\[tan15{}^\circ \text{ }+\text{ }tan30{}^\circ \text{ }+\text{ }tan15{}^\circ \text{ }tan30{}^\circ \text{ }=\text{ }1\]
Hence proved.