Proving trig identities: sec x + csc x = (1 + tan x)/sin x

[anna_sims]

I need some guidance on proving these identitles. Any help would be appreciated.

1)

((sec x+1)(sec x-1))/(2+(tan x+1)(tan x-1)) = sin^2 x

2)

sec x + csc x = (1+tan x/sin x)

 

[arthur ohlsten]

2)
secx + cscx= [1+tanx]/sinx or
1/cosx + 1/sinx = [1+sinx/cosx] / sinx
multiply both sides by sinx
sinx/cosx +1 =1+ sinx/cosx proof

1)
[secx+1][secx-1]
--------------------- = sin^2x
2+[tanx+1][tanx-1]

[ 1/cosx+1][1/cosx -1]
------------------------------------- = sin^2x
2+ [sinx/cosx +1][sinx/cosx -1]

[ 1+cosx][1-cosx] / cos^2x
---------------------------------------- = sin^2x
2+[sinx+cosx][sinx-cosx]/cos^2x

multiply numerator and denominator by cos^2x x not =90,270 degrees

[1+cosx][1-cosx]
------------------------------------- = sin^2x
2cos^2x+[sinx+cosx][sinx-cosx]

[1-cos^2x]
------------------------------------- = sin^2x
2 cos^2x+sin^2x-cos^2x

sin^2x / [sin^2x+cos^2x] = sin^2x

sin^2x=sin^2x proof

Arthur