Question

pls prove the below question
tanx-sinx/sin^2x=Tanz/1+cosx
​

Answer 1

Step-by-step explanation:

We have,

$\frac{ \tan(x) - \sin(x) }{ \sin ^{2} (x) }$

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

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

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

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

$= \frac{ \tan(x) }{ 1 + \cos(x) }$