Answer:
1) [tex]\frac{\sin x}{1-\cos x} = \csc x + \cot x[/tex]
2) [tex]\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}[/tex]
3) [tex]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}[/tex]
4) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}[/tex]
5) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}[/tex]
6) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}[/tex]
7) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}[/tex]
Step-by-step explanation:
Now we proceed to show all steps needed to demonstrate the trigonometric identity:
1) [tex]\frac{\sin x}{1-\cos x} = \csc x + \cot x[/tex] Given.
2) [tex]\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}[/tex] Identities for cosecant and cotangent functions.
3) [tex]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}[/tex] [tex]\frac{a}{b}+\frac{c}{b} = \frac{a+c}{b}[/tex]
4) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}[/tex] Existence of additive inverse/Modulative property.
5) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}[/tex] Fundamental trigonometric identity.
6) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}[/tex] Factorization.
7) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}[/tex] Existence of additive inverse/Modulative property/Result.
