There is no solution, and none is required, because there's no question here yet.
There's no equation.
There is only a trigonometric expression, whose numerical value changes every time
'x' changes.
If we were told that the expression equals something, then we would have have an
equation, which could be solved for one or more 'x' values that satisfy it.
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d/dx(1+tanx)=0+sec2x=sec2x
(-x+tanx)'=-1+(1/cos2x)
I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)