Remembering the derivative of #arctanx# here is very helpful. If you don't, it could get a bit frustrating with implicit differentiation. I'll do it both ways to show you what I mean.

#d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))#

Thus, we have:

#y = arctan(secx + tanx)#

#(dy)/(dx) = 1/(1+(secx + tanx)^2)(secxtanx + sec^2x)#

#= (secxtanx + sec^2x)/(1+(secx + tanx)^2)#

#= (secxtanx + sec^2x)/(1+sec^2x + 2secxtanx + tan^2x)#

#= (sec^2x + secxtanx)/(2sec^2x + 2secxtanx)#

#= color(blue)(1/2)#

(Haha, nice. A derivative that doesn't even integrate back into the original function without some special manipulation.)

And now the other way.

#tany = secx + tanx#

#sec^2y * (dy)/(dx) = secxtanx + sec^2x#

#(1 + tan^2color(green)(y)) * (dy)/(dx) = secxtanx + sec^2x#

Note that you have to use this identity, otherwise you'll get #sec^2(arctan(secx + tanx))#, which is not easy to work with.

#[1 + (tan[color(green)(arctan(secx + tanx))])^2] (dy)/(dx) = secxtanx + sec^2x#

#[1 + (secx + tanx)^2] (dy)/(dx) = secxtanx + sec^2x#

#(dy)/(dx) = (secxtanx + sec^2x)/(1 + (secx + tanx)^2)#

which, from above, is:

#color(blue)(= 1/2)#