Q: Why the values of tanx is pi?

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pi radians.

1 because tan(5 pi / 4) = 1

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)

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pi radians.

1 because tan(5 pi / 4) = 1

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)

Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|

I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1

we do not check if a function is continuous or not outside it's domain."first, f has to be defined at c."Tanx is not defined where cosx=0 .ie x=pi/2 , 3pi/2 etcill try to help more here.what domain means is what can you put into a function, whereas range, which i am sure you have heard of as well, just means what you can get out of a function. that being said, lets look further into the graph of tanx. when we do, we see that the graph is discontinuous at pi/2. the reason for this is because tanx is equivalent to sinx/cosx. because of this relationship, when you put pi/2 in for x in sinx/cosx, you end up with cosx=0 which makes your denominator zero, which is undefined, which makes your graph discontinuous. because of that, you cannot put pi/2 in for x in tanx, and since the domain is what you can put into an equation, pi/2 which causes a discontinuity is not included in the domain. basicly, wherever a graph is discontinuous, it wont be included in the domain because you cant put stuff in that will make your graph discontinuous

d/dx(1+tanx)=0+sec2x=sec2x

Pi cannot be expressed exactly as any fraction (including as a fraction of powers of 10, which is what a decimal fraction is). There are an infinite number of place values in the number 'pi'.

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)

you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx