(i)
(ii)
Solution:
(i)
\[=~tan\text{ }A\]
\[=\text{ }RHS\]
\[\therefore LHS\text{ }=\text{ }RHS\]
Hence proved.
(ii)
\[=~tan\text{ }A\text{ }-\text{ }tan\text{ }B\text{}+\text{}tan\text{}B\text{}-\text{}tan\text{}C\text{}+\text{}tan\text{}C\text{}-\text{}tan\text{}A\]
\[=\text{ }0\]
\[=\text{ }RHS\]
\[\therefore LHS\text{ }=\text{ }RHS\]
Hence proved.