Precal/Trig
posted by Katherine .
(tan/(1+sec)) + ((1+sec)/tan) = 2csc
(Show all work please.)

I see tangents and secants on the left side, so I suspect to use the identity
tan^2x + 1 = sec^2x
LS
= tanx(1+secx) + (1+secx)/tanx
= (tan^2x + 1 + 2secx + sec^2x)/(tanx(1+secx))
= (sec^2x + 2secx + sec^2x)/(tanx(1+secx))
= (2sec^2x + 2secx)/(tanx(1+secx))
= 2secx(secx + 1)/(tanx(1+secx))
= 2secx/tanx
= (2/cosx)÷(sinx/cosx)
= (2/cosx)(cosx/sinx)
= 2/sinx
= 2cscx
= RS
Q.E.D. 
y=2sec(1/3x)
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