sec^2(x)
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y' = (sec(x))^2
Tanx was created in 1972-10.
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
(-x+tanx)'=-1+(1/cos2x)
I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1