By considering \(\sum _ { k = 0 } ^ { n - 1 } ( 1 + \mathrm { i } \tan \theta ) ^ { k }\), show that $$\sum _ { k = 0 } ^ { n - 1 } \cos k \theta \sec ^ { k } \theta = \cot \theta \sin n \theta \sec ^ { n } \theta$$ provided \(\theta\) is not an integer multiple of \(\frac { 1 } { 2 } \pi\). Hence or otherwise show that $$\sum _ { k = 0 } ^ { n - 1 } 2 ^ { k } \cos \left( \frac { 1 } { 3 } k \pi \right) = \frac { 2 ^ { n } } { \sqrt { 3 } } \sin \left( \frac { 1 } { 3 } n \pi \right)$$ Given that \(0 < x < 1\), show that $$\sum _ { k = 0 } ^ { n - 1 } \frac { \cos \left( k \cos ^ { - 1 } x \right) } { x ^ { k } } = \frac { \sin \left( n \cos ^ { - 1 } x \right) } { x ^ { n - 1 } \sqrt { } \left( 1 - x ^ { 2 } \right) }$$