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.
It is -sqrt(1 + cot^2 theta)
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
d/dx (-cscx-sinx)=cscxcotx-cosx
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
The derivative of csc(x) is -cot(x)csc(x).
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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.
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
Yes.
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
csc^2x+cot^2x=1
yes 1 + cot x^2 = csc x^2
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
d/dx csc(x) = - csc(x) tan(x)
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
Without an "equals" sign somewhere, no question has been asked,so there's nothing there that needs an answer.Is it the sum that you're looking for ?csc(x) + cot(x) = 1/sin(x) + cos(x)/sin(x) = [1 + cos(x)] / sin(x)