Prove the following  $\tan \theta +\tan \left(90°-\theta \right)= \sec \theta \sec \left(90°-\theta \right)$

Simplify $\left(1+{\tan }^2 \theta \right) \left(1-\sin \theta \right) \left(1+ \sin \theta \right)$.

Show that $\frac{{\cos }^2\left(45°+\theta \right)+{\cos }^2\left(45°-\theta \right)}{\tan \left(60°+\theta \right) \tan \left(30°-\theta \right)}=1$

An observer 1.5 m tall is 20.5 m away from a tower  22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.