1+sin 2x/cos x +sin x = cos 2x/cos x - sin x?

Question

Verify this Identity

Answer 1

Your problem is not clear enough. There may be several possibilities .
Possibility 1 :

LHS= 1 + [sin(2*x)/cos(x)] + sin(x)
Use sin(2*x) = 2* sin(x) * cos(x)
Hence,
LHS= 1 + 2* sin(x) + sin(x)
= 1+ 3*sin(x)

RHS = [cos(2*x)/cos(x)] - sin(x)
Use cos(2*x) = cos^2(x) - sin^2(x)
= 1- sin^2(x) - sin^2(x)
= 1 - 2*sin^2(x)
Hence,
RHS = [1 - 2* sin^2(x) ]/cos(x) - sin(x)
= [1- 2* sin^2(x) - sin(x) * cos(x) ]/cos(x)
= 2*cos(x)-(1/cos(x))
-sin(x)

which is not equal to the LHS.

The problem is wrong and it is not an identity.

Possibility 2:
LHS= (1 + sin(2*x))/(sin(x) + cos(x))
=[ 2 * sin(x) * cos(x) + cos^2(x) + sin^2(x)]
* 1/[cos(x) + sin(x) ]
[using 1 = cos^2(x) + sin^2(x) ]
= [cos(x) + sin(x)]^2
* 1/[cos(x) + sin(x) ]
= cos(x) + sin(x)

RHS = cos(2*x)/[cos(x) -sin(x)]
= [cos^2(x) - sin^2(x)]
* 1/[cos(x) - sin(x)]
= {cos(x) + sin(x) }
* {cos(x) - sin(x)}
* 1/[cos(x) - sin(x)]
[using a^2 -b^2 = (a+b)*(a-b)]

= cos(x) + sin(x)

Hence LHS = RHS

Proved.

Answer 2

You must show groupings with parentheses, otherwise few people will understand what you are asking.

We will use the following double angle formulas:
sin 2x = 2 sin x cos x
cos 2x = cos²(x) - sin²(x)

LHS (Left hand side) =
(1+sin(2x))/(cos(x) + sin(x)) =
(cos²(x) + sin²(x) + 2sin(x)cos(x))/(cos(x)+sin(x)) =
(cos(x)+sin(x))² / (cos(x)+sin(x)) =
cos(x) + sin(x) =
(cos(x)+sin(x)) (cos(x)-sin(x)) / (cos(x)-sin(x)) =
(cos²(x)-sin²(x)) / (cos(x)-sin(x)) =
cos(2x)/(cos(x)-sin(x)) =
RHS