+0

# question

0
57
4
+1898

Apr 22, 2020

#1
+928
+1

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

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

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

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

.
Apr 22, 2020
#2
+928
+1

Sorry this took so long... I was missing a step and it took me FOREVER to find out what it was.

HELPMEEEEEEEEEEEEE  Apr 22, 2020
#3
+1898
+1

no its okay , thank you so much for yuor help. gonna look over it right now and see if  i have any questions

jjennylove  Apr 22, 2020
#4
+24883
+2

Verify the identity:
$$\dfrac{\sin(x)}{1-\cos(x)} = \csc(x)+\cot(x)$$

$$\begin{array}{|rcll|} \hline \dfrac{\sin(x)}{1-\cos(x)} &=& \csc(x)+\cot(x) \quad & | \quad \csc(x) = \dfrac{1}{\sin(x)} \\\\ \dfrac{\sin(x)}{1-\cos(x)} &=& \dfrac{1}{\sin(x)}+\cot(x) \quad & | \quad \cot(x) = \dfrac{\cos(x)}{\sin(x)} \\\\ \dfrac{\sin(x)}{1-\cos(x)} &=& \dfrac{1}{\sin(x)}+\dfrac{\cos(x)}{\sin(x)} \\\\ \dfrac{\sin(x)}{1-\cos(x)} &=& \dfrac{1+\cos(x)}{\sin(x)}\quad & | \quad \times \sin(x) \\\\ \dfrac{\sin^2(x)}{1-\cos(x)} &=& 1+\cos(x) \quad & | \quad \times \Big(1-\cos(x)\Big) \\\\ \sin^2(x) &=&\Big(1+\cos(x)\Big)\Big(1-\cos(x)\Big) \\\\ \sin^2(x) &=& 1-\cos^2(x) \quad & | \quad + \cos^2(x) \\\\ \sin^2(x)+ \cos^2(x) &=& 1 \ \checkmark \qquad \text{Pythagorean Theorem} \\ \hline \end{array}$$

Apr 23, 2020