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to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities..
Cosx + Sinx Tanx
Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx
Cosx + Sin2x/ Cos x <------------- do the LCD
Cosx (Cosx/Cosx) + Sin2x/Cosx
(Cos2x+Sin2x)/Cosx
1/Cosx <--------- From Sin2x + Cos2x =1
or Secx <-------- answer
Comment if you have questions...:))
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Cos x plus sin x equals 0?
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
Does sin x tan x plus cos x equals tan x?
NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.
What is the limit of x- sin x cos x over tan x -x as x tends to zero?
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1
Why tan x graph is continuous within its domain and is not continuous out of the domain?
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
How do you differentiate 2tanx?
Coefficients can be removed when differentiating, i.e. d/dx 2tanx = 2 d/dx tanx You should know that d/dx tanx = sec2x (either by differentiating sinx/cosx or by just remembering the derivatives of common trig functions, as it will come in handy). So the answer is 2sec2x
How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
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)
Can you Show 1 over sinx cosx - cosx over sinx equals tanx?
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
Tan plus cot divided by tan equals csc squared?
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
How do you Prove sin x times sec x equals tan x?
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
Cos x plus sin x equals 0?
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
How do you prove the following identity sec x - cos x equals sin x tan x?
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
Does sin x tan x plus cos x equals tan x?
NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.
How do you simplify tanx plus tany per cosx plus coty?
(tan x- 1)/ (1+tan x)
How do you solve csc x-sin x equals cos x cot x?
cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.
Cos x tan x to sin x Transform the expression on the left to the one on the right?
cosx * tanx = sinx cos x . sin x/ cosx = sinx cos x . sin x/cosx = sinx (cos x is taken out of the equation, leaving sin x = sin x).
When are y equals sin x and y equals cos x equal?
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
