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The proof of this trig identity relies on the pythagorean trig identity, the most famous trig identity of all time: sin2x + cos2x = 1, or 1 - cos2x = sin2x.
1 + cot2x = csc2x
1 = csc2x - cot2x
1 = 1/sin2x - cos2x/sin2x
1 = (1 - cos2x)/sin2x ...using the pythagorean trig identity...
1 = sin2x/sin2x
1 = 1
So this is less of a proof and more of a verification.
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What is the anti-derivative of secant squared?
negative cotangent -- dcot(x)/dx=-csc^2(x)
What is the range of csc x?
It is R excluding (-1, 1) if x is real.
Definition of the 6 trigonometric functions?
The six main trigonometric functions are sin(x)=opposite/hypotenuse cos(x)=adjacent/hypotenuse tan(x)=opposite/adjacent csc(x)=hypotenuse/opposite cot(x)=adjacent/opposite sec(x)=hypotenuse/adjacent Where hypotenuse, opposite, and adjacent correspond to the three sides of a right triangle and x corresponds to an angle in that right triangle.
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
Which of the six trigonometric functions are even functions?
A function f(x) is even if: f(x) = f(-x) In layman's terms this property simply means that any real number in the domain and it's opposite will yield the same function value in the range. To simplify this down even further, an even function, when graphed will appear to be symetric about the y-axis (assuming that you use the standard Cartesian coordinate plane). In the case of trig functions, you would have to test whether the even function property holds true for each. We will the test points π, π/2, or π/4. NOTE: The # signs are present next to the functions that are even: 1. Sine: f(x) = sin(x) -> sin(π/2) = 1, but sin(-π/2) = -1. Since 1 does not equal -1, sine is NOT an even function. 2. #Cosine: f(x) = cos(x) -> cos(π) = -1 = cos(-π). Since both are equal, cosine IS an even function. 3. Tangent: f(x) = tan(x) -> tan(π/4) = 1, but tan(-π/4) = -1. Therefore, tangent is NOT an even function. 4. Cosecant: f(x) = csc(x) -> csc(π/2) = 1, but csc(-π/2) = -1. Therefore, cosecant is NOT an even function. 5. #Secant: f(x) = sec(x) -> sec(π) = -1 = sec(-π). Since the secant function has asymptotes, it IS an even function provided that x does not equal π(2n+1)/2, where n may be all integers. 6. Cotangent: f(x) = cot(x) -> cot(π/4) = 1, but cot(-π/2) = -1. Therefore cotangent is NOT even.
Cotangent squared plus one equals cosecant squared?
yes 1 + cot x^2 = csc x^2
Csc divided by cot squared equals tan multiplied by sec?
Yes.
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.
Csc squared divided by cot equals csc x sec. can someone make them equal?
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)
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
What is co-tangent squared?
Start with the identity (sin a)2 + (cos a)2 = 1. Divide both sides by (sin a)2 to get1 + (cot a)2 = (csc a)2. Then subtract 1 from both sides. (cot a)2 = (csc a)2 - 1.
What is the answer of csc x plus cot 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)
Can you simplify 1-cot x?
csc^2x+cot^2x=1
What is the anti derivative of cscxcotx?
∫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
How do you solve csc x sin x equals cos x cot x plus?
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)
How do you integrate the cscnx?
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
What is sec squared x times csc x divided by sec squared x plus csc squared x?
Ah, secant, annoying as always. Why don't we use its definition as 1/cos x and csc as 1/sin x? We will do that Also, please write down the equation, there is at least TWO different equations you are talking about. x^n means x to the power of n 1/(sin x) ^2 is csc squared x, it's actually csc x all squared 1/(cos x) ^2 in the same manner.
