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avatar+1898 

i dont even know where to begin , can someone please help.

 Apr 22, 2020
 #1
avatar+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
avatar+928 
+1

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

HELPMEEEEEEEEEEEEE  Apr 22, 2020
 #3
avatar+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
avatar+24883 
+2

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

 

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

 

laugh

 Apr 23, 2020

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