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No, they are the inverse functions, while csc, sec and cot are the reciprocal functions.

To illustrate the difference, the inverse of f(x) = x+3 is f-1(x) = x-3

But the reciprocal of f(x) is 1/f(x) = 1/(x+3)

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Q: Are shifts sin cos tan equal to csc sec cot respectively?
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How do you simplify cos times cot plus sin?

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec


Does cotangent plus one equal cosecant?

Cotangent = 1/Tangent : Cosecant = 1/Sine Then, cot + 1 = (1/tan) + 1 = (cos/sin) + (sin/sin) = (cos + sin)/ sin. Now, cos² + sin² = 1 so for the statement to be valid the final expression would have to be : (cos² + sin² ) / sin = 1/sin. As this is not the case then, cot + 1 ≠ cosec. In fact, the relationship link is cot² + 1 = cosec²


How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?

Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.


Is 2 cot x sin x cos x equals 2 - 2 sin 2 x an identity?

The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an 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)

Related questions

Verify cot x-180 cot x?

cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x


How do you simplify csc theta cot theta cos theta?

cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)


How do you simplify cos times cot plus sin?

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec


Does cotangent plus one equal cosecant?

Cotangent = 1/Tangent : Cosecant = 1/Sine Then, cot + 1 = (1/tan) + 1 = (cos/sin) + (sin/sin) = (cos + sin)/ sin. Now, cos² + sin² = 1 so for the statement to be valid the final expression would have to be : (cos² + sin² ) / sin = 1/sin. As this is not the case then, cot + 1 ≠ cosec. In fact, the relationship link is cot² + 1 = cosec²


When does cotangent equal -1?

cot[x]= -1 cot[x] = cos[x] / sin[x] cos[x] / sin[x] = -1 cos[x] = -sin[x] |cos[x]| = |sin[x]| at every multiple of Pi/4 + Pi/2. However, the signs disagree at 3Pi/4 + nPi, where n is an integer.


How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?

Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.


Is 2 cot x sin x cos x equals 2 - 2 sin 2 x an identity?

The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an 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)


Simplify sinx cotx cosx?

== cot(x)== 1/tan(x) = cos(x)/sin(x) Now substitute cos(x)/sin(x) into the expression, in place of cot(x) So now: sin(x) cot(x) cos(x) = sin(x) cos(x) (cos(x)/sin(x) ) sin(x) cos(x) cos(x)/sin(x) The two sin(x) cancel, leaving you with cos(x) cos(x) Which is the same as cos2(x) So: sin(x) cot(x) cos(x) = cos2(x) ===


What is the cot of a 68 degree angle?

The trig identaty of cot(x) is cos(x)/sin(x) so then if we want to evaluate cot (68) deg. we just plug into the identady. so cos(68)/sin(68)=.404


What are the different types of number system in scientific calculator?

tan cot sec cosec sin cos cot


How do you simplify sec x cot x cos x?

y = sec(x)*cot(x)*cos(x)To solve this trigonometric equation, you need to know these identities:sec(x) = 1/(cos(x))cot(x) = 1/(tan(x)) = (cos(x))/(sin(x))Now substitute these identities into the original equation:y = (1/cos(x))*((cos(x))/(sin(x)))*cos(x)Now cancel out the terms that are similar in the numerator and denominator to leave you with:y = (1/(sin(x)))*cos(x)y = (cos(x))/(sin(x))From the aforementioned known identity, the final simplified trigonometric equation becomes:y = cot(x)