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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)
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
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 can you prove that 1-2 cosine squared over sine times cosine is equal to tangent minus cotangent?
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
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)
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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)
1 over tan x equals what?
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
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
Cos2x equals 1 minus tan squared x divide by1 plus tan squared x?
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
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.
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) ===
