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The statement "cot multiplied by cosec equals cos" is not accurate. In trigonometric terms, cotangent (cot) is the reciprocal of tangent, and cosecant (cosec) is the reciprocal of sine. Therefore, the correct relationship is ( \cot(x) \cdot \csc(x) = \frac{\cos(x)}{\sin^2(x)} ), which does not simplify to cosine. Instead, it highlights the relationship between these functions in terms of sine and cosine.
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What are the different types of number system in scientific calculator?
tan cot sec cosec sin cos cot
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 calculate sec in trigonometry?
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
How do you simplify tan cot equals 1?
It just simplifies down to 1=1. You have to use your trig identities... tan=sin/cos cot=cos/sin thus tan x cot= (sin/cos) (cos/sin) since sin is in the numerator for tan, when it is multiplied by cot (which has sin in the denominator) both of the signs cancel and both now have a value of 1. The same happens with cos. so you get 1 x 1=1 so there is your answer. just learn your trig identities and you will understand
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)
What are the different types of number system in scientific calculator?
tan cot sec cosec sin cos cot
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
What are the different types of number systems available in scientific calculator?
tan cot sec cosec sin cos cot
Second derivative of cosecx?
d/dx cosec(x) = - cosec(x) * cot(x) so the second derivative or d(d/dx)/dx cosec(x) = [- cosec(x) * d/dx cot(x)] + [ - d/dx cosec(x) * cot(x)] = [- cosec(x) * -cosec^2(x)] + [ - (- cosec(x) * cot(x)) * cot(x)] = cosec(x) * cosec^2(x) + cosec(x)*cot^2(x) = cosec(x) * [cosec^2(x) + cot^2(x)].
What is cot x?
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
Which functions have a period of Pi?
sin(2x), cos(2x), cosec(2x), sec(2x), tan(x) and cot(x) are all possible functions.
What are different types of number systems available in scientific calculator?
Sin Cos Tan Sec Cosec Cot
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
What are the 6 trig functions for 0 degrees 90 degrees 180 degrees 270 degrees?
sin(0) = 0, sin(90) = 1, sin(180) = 0, sin (270) = -1 cos(0) = 1, cos(90) = 0, cos(180) = -1, cos (270) = 0 tan(0) = 0, tan (180) = 0. cosec(90) = 1, cosec(270) = -1 sec(0) = 1, sec(180) = -1 cot(90)= 0, cot(270) = 0 The rest of them: tan(90), tan (270) cosec(0), cosec(180) sec(90), sec(270) cot(0), cot(180) are not defined since they entail division by zero.
How do you calculate sec in trigonometry?
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
How do you simplify tan cot equals 1?
It just simplifies down to 1=1. You have to use your trig identities... tan=sin/cos cot=cos/sin thus tan x cot= (sin/cos) (cos/sin) since sin is in the numerator for tan, when it is multiplied by cot (which has sin in the denominator) both of the signs cancel and both now have a value of 1. The same happens with cos. so you get 1 x 1=1 so there is your answer. just learn your trig identities and you will understand
