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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)
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
What is the integral of underroot tanx plus underroot cotxdx?
It isn't clear what you mean by "underroot"; I am not aware of a mathematical function by that name.
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
How do you solve 6tanx2-17tanx plus 7 equals 0?
Assuming your equation is: 6(tanx)2-17tanx + 7 = 0The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:6s2-17s+7=0Using the quadratic equation solve for s:s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12s = 28/12 or 7/3 & s = 6/12 or 1/2Now substitute tanx back for s to gettan x = 7/3 & tan x = 1/2x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6oIn radians it would be 1.166 & 0.4636
What is the integral of tan cubed x secx dx?
This is a trigonometric integration using trig identities. S tanX^3 secX dX S tanX^2 secX tanX dX S (secX^2 -1) secX tanX dX u = secX du = secX tanX S ( u^2 - 1) du 1/3secX^3 - secX + C
Intergrate sec x?
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
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
Integration by parts of x tanx?
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
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)
Integration of root tanx?
The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on Answers.com. See the related links for a representation of the answer.
How do you integrate sec cube 2x?
The integral of sec(x) is ln|secx+tanx| + C Since the derivative is taken to the third power, we have to consider the chain rule; the original equation must be to the fourth power, and in order for that to be canceled out, the equation must also have had a coefficient of 1/4. 2x is also subject to the chain rule. I would suggest u substitution. integral(sec(2x))^3 dx u=2x du=2dx dx=1/2du integral (sec(u))^3 *1/2 du 1/8 secxtanx + 1/8(ln|secx+tanx|^4) + C
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
Integration of tan pow4x?
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
What is the derivative of 5secx?
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
How do you simplify cosx plus sinx tanx?
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...:))
What is the integral of underroot tanx plus underroot cotxdx?
It isn't clear what you mean by "underroot"; I am not aware of a mathematical function by that name.
