Question

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Answer 1

Answer:

1) $\frac{\sin x}{1-\cos x} = \csc x + \cot x$

2) $\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}$

3) $\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$

4) $\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}$

5) $\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}$

6) $\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}$

7) $\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}$

Step-by-step explanation:

Now we proceed to show all steps needed to demonstrate the trigonometric identity:

1) $\frac{\sin x}{1-\cos x} = \csc x + \cot x$ Given.

2) $\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}$ Identities for cosecant and cotangent functions.

3) $\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$             $\frac{a}{b}+\frac{c}{b} = \frac{a+c}{b}$

4) $\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}$ Existence of additive inverse/Modulative property.

5) $\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}$    Fundamental trigonometric identity.

6) $\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}$   Factorization.

7) $\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}$ Existence of additive inverse/Modulative property/Result.