Answer:
The answer in the procedure
Step-by-step explanation:
we have
[tex]\frac{sin(x)}{1-cos(x)}+\frac{sin(x)}{1-cos(x)}=2csc(x)[/tex]
Prove the identity
In this problem there is a mistake: in order to obtain an identity the second denominator at the left side must be 1 + cosx
so
[tex]\frac{sin(x)}{1-cos(x)}+\frac{sin(x)}{1+cos(x)}=2csc(x)[/tex]
Adds fraction in the left side
[tex]\frac{sin(x)(1+cos(x)+sin(x)(1-cos(x)}{1-cos^{2} (x)}=2csc(x)\\ \\\frac{2sin(x)}{1-cos^{2} (x)}=2csc(x)[/tex]
Remember that
[tex]csc(x)=\frac{1}{sin(x)}[/tex]
and
[tex]sin^{2}(x)+cos^{2}(x)=1[/tex]
[tex]sin^{2}(x)=1-cos^{2}(x)[/tex]
substitute
[tex]\frac{2sin(x)}{sin^{2}(x)}=2\frac{1}{sin(x)}[/tex]
Multiply both sides by sin(x)
[tex](sin(x))\frac{2sin(x)}{sin^{2}(x)}=2[/tex]
[tex]\frac{2sin^{2}(x)}{sin^{2}(x)}=2[/tex]
[tex]2=2[/tex] ----> identity verified
