Prove the identity \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 \equiv \frac{1 - \cos x}{1 + \cos x}\).
Prove the identity \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} \equiv \frac{\tan \theta - 1}{\tan \theta + 1}\).
Show that \(\sin^4 \theta - \cos^4 \theta \equiv 2 \sin^2 \theta - 1\).
Prove the identity \(\frac{\tan x + 1}{\sin x \tan x + \cos x} \equiv \sin x + \cos x\).
Prove the identity \(\frac{1}{\cos \theta} - \frac{\cos \theta}{1 + \sin \theta} \equiv \tan \theta\).
Prove the identity \(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} \equiv \frac{1}{\tan \theta}\).
Show that \(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta} \equiv \frac{1}{\sin^2 \theta - \cos^2 \theta}\).
Prove the identity \(\tan^2 \theta - \sin^2 \theta \equiv \tan^2 \theta \sin^2 \theta\).
Show that \(\frac{\tan \theta}{\sin \theta} - \frac{\sin \theta}{\tan \theta} \equiv \tan \theta \sin \theta\).
Prove the identity \(\tan x + \frac{1}{\tan x} = \frac{1}{\sin x \cos x}\).
Prove the identity \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 \equiv \frac{1 - \cos \theta}{1 + \cos \theta}\).
Prove the identity \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} \equiv 1 + \frac{1}{\sin \theta}\).
Prove the identity
\(\tan^2 x - \sin^2 x \equiv \tan^2 x \sin^2 x\).
Prove the identity \(\frac{\sin x \tan x}{1 - \cos x} \equiv 1 + \frac{1}{\cos x}\).
Prove the identity \((\sin x + \cos x)(1 - \sin x \cos x) \equiv \sin^3 x + \cos^3 x\).
Prove the identity \(\frac{\sin x}{1 - \sin x} - \frac{\sin x}{1 + \sin x} \equiv 2 \tan^2 x\).
Prove the identity
\(\frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} = \frac{2}{\cos x}.\)
Prove the identity \(\frac{1 - \tan^2 x}{1 + \tan^2 x} \equiv 1 - 2 \sin^2 x\).
Prove the identity \(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta} \equiv \frac{\tan^2 \theta + 1}{\tan^2 \theta - 1}\).
Prove the identity \(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} \equiv -\tan^2 \theta (1 + \sin^2 \theta)\).