(a)[4]
Prove that $\left(\frac{1}{\cos x} - \tan x\right)\left(\frac{1}{\sin x} + 1\right) = \frac{1}{\tan x}$.
(b)[2]
Hence solve $\left(\frac{1}{\cos x} - \tan x\right)\left(\frac{1}{\sin x} + 1\right) = 2\tan^2 x$ for $0^\circ \leq x \leq 180^\circ$.