- (a) Use \(t = \tan \frac { \theta } { 2 }\) to show that, where both sides are defined
$$\frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta } \equiv \frac { 5 t + 2 } { 2 t + 5 }$$
(b) Hence, again using \(t = \tan \frac { \theta } { 2 }\), prove that, where both sides are defined
$$\frac { 20 + 21 \tan \theta } { 29 + 21 \sec \theta } \equiv \frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta }$$