Question

(1+cosA)/SinA=SinA/(1-CotA)

Answer 1

hey it is impossible plz check your question

Answer 2

Your question needs a correction :

cotA will be replaced with cosA ,




Solution :

LHS,
$\frac{1 + \cos(a) }{ \sin(a) }$


Multiply and divide by ( 1 - cos A ) ,


$= > \frac{(1 + \cos(a))(1 - \cos(a) ) }{ \sin(a)(1 - \cos(a) ) } \\ \\ \\ \\ = > \frac{ {1}^{2} - {cos}^{2} (a)}{ \sin(a)(1 - \cos(a) ) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ | \: \: (a + b)(a - b) = {a}^{2} - {b}^{2} } \\ \\ \\ \\ = > \frac{1 - {cos}^{2} (a)}{ \sin(a)(1 - \cos( a) ) } \\ \\ \\ \\ = > \frac{ \sin^{2} (a) }{ \sin(a) ( 1 - \cos(a) )} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ | 1 - {cos}^{2} (a) = {sin}^{2} (a)}\\ \\ \\ \\ = > \frac{sin(a)}{1 - cos(a)}$



RHS,


$\frac{ \sin(a) }{1 - \cos(a) }$






LHS = RHS



Hence, proved.