Question

verify: 1-cosx/six+sinx/1-cosx=2cscx with steps please i have an exam tomorrow

Answer 1

Try combining the fractions.

Answer 2

You could use the following identities to simplify the left-hand side:
\[\frac ab + \frac cd = \frac{ad + bc}{bd}\\\cos^2(x) + \sin^2(x) = 1\]

Answer 3

Doing so you would get
\[\frac{1-\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 - \cos(x)} = \frac{(1 - \cos(x))^2 + \sin^2(x)}{\sin(x)(1-\cos(x))}\\ = \frac{1 - 2\cos(x) + \cos^2(x) + \sin^2(x)}{\sin(x)(1-\cos(x))}\\= \frac{1 - 2\cos(x) + 1}{\sin(x)(1-\cos(x))}\\= \frac{2(1 - \cos(x))}{\sin(x)(1-\cos(x))}\\=\frac 2{\sin(x)}=2\csc(x)\]

Answer 4

thanks man :D life saver but how did you get the 2cosx ?