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Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
Wiki User
∙ 13y ago
(1 + tanx)/sinx
Multiply by sinx/sinx
sinx + tanxsinx
Divide by sin2x (1/sin2x) = cscx
cscx + tan(x)csc(x)
tanx = sinx/cosx and cscx = 1/sinx
cscx + (sinx/cosx)(1/sinx)
sinx cancels out
cscx + 1/cosx
1/cosx = secx
cscx + secx
Wiki User
∙ 13y ago
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Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
How do you verify the identity sinx cscx 1?
sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
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 rewrite y equals cscx to graph it?
You could try y = 1/sin(x) but I do not see how that helps.
Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
What is the integral of cscx?
- ln (cscx + cotx) + C You use u substitution.
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
How do you prove that the derivative of csc x is equals to -csc x cot x?
d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.
How do you verify the identity sinx cscx 1?
sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
How do you find the derivative of - csc x - sin x?
d/dx (-cscx-sinx)=cscxcotx-cosx
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
What is the derivative of -cscxcotx?
Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.
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.
What is the anti derivative of the square root of 1-x2?
-1
