That is not correct. "tan x" is a function that depends on the value of "x"; it is not always pi.
tan(x) is pi only if 'x' is about 72.3 degrees, 252.3 degrees, or either of these added to a multiple of 360 degrees.
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)
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
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'.
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
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
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)