There is insufficient information because of undisplayed text, so I’m guessing what the identity is meant to be.
tan(x)/(1+cos(x))+tan(x)/(1-cos(x))=
(sin(x)/cos(x))(1/(1+cos(x))+1/(1-cos(x))=
(sin(x)/cos(x))(1-cos(x)+1+cos(x))/(1-cos²(x))=
(sin(x)/cos(x))(2/sin²(x))=
2/(sin(x)cos(x)).
So the identity is:
tan(x)/(1+cos(x))+tan(x)/(1-cos(x))≡2/(sin(x)cos(x)).
I’m also guessing the second part of the question:
tan(x)/(1+cos(x))+tan(x)/(1-cos(x))=6tan(x) or 6/tan(x).
Assume 6/tan(x):
2/sin(x)cos(x)=6/tan(x), 1/sin(x)cos(x)=3/tan(x),
sin(x)cos(x)=sin(x)/(3cos(x)),
3sin(x)cos²(x)-sin(x)=0,
sin(x)(3cos²(x)-1)=0,
sin(x)=0⇒x=0°, 180° or 360°; or:
cos(x)=±√3/3, x=54.74°, 125.26°, 234.74°, 305.26° if x is between 0° and 360°.
Assume 6tan(x):
2/sin(x)cos(x)=6sin(x)/cos(x),
(1/cos(x))(1/sin(x)-3sin(x))=0.
Since 1/cos(x) cannot be zero:
3sin²(x)=1, sin(x)=±√3/3, x=35.26°, 144.74°, 215.26°, 324.74° for 0°≤x≤360°.