Question

Sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

Answer 1

Answer:

The answer in the procedure

Step-by-step explanation:

we have

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

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

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

Adds fraction in the left side

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

Remember that

$csc(x)=\frac{1}{sin(x)}$

and

$sin^{2}(x)+cos^{2}(x)=1$

$sin^{2}(x)=1-cos^{2}(x)$

substitute

$\frac{2sin(x)}{sin^{2}(x)}=2\frac{1}{sin(x)}$

Multiply both sides by sin(x)

$(sin(x))\frac{2sin(x)}{sin^{2}(x)}=2$

$\frac{2sin^{2}(x)}{sin^{2}(x)}=2$

$2=2$ ----> identity verified