Question

sinA/1-cosA=cosecA+cotA​

Answer 1

hope this answer will be helpful ^_^

Answer 2

Let's do it,

$\frac{ \sin \: A }{1 - \cos \: A} = \csc \: A + \cot \: A$

Using LHS:

multiplying numerator and denominator by (1 + cos A)

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

Since we have an identity as followed:
${ \sin }^{2} \alpha + { \cos }^{2} \alpha = 1 \\ \\1 - { \cos}^{2} \alpha = { \sin }^{2} \alpha$
So using this we have,
$\frac{ \sin \: A(1 + \cos \: A )}{ { \sin }^{2}A} \\ \\ = \frac{1 + \cos \: A}{ \sin \: A} \\ \\ = \frac{1}{ \cos \: A } + \frac{ \cos \: A }{ \sin \: A } \\ \\ = \csc A + \cot \: A \: \\ \\ = RHS$


Hence proved.

Hope it was satisfying enough.