If Cosec K-Cot K = 0. find sin K
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Answers
Step-by-step explanation:
What is the solution of cot (x) +cosec(x) =0?
If cotx+cscx=0 , then multiplying both sides by cscx−cotx , we get csc2x−cot2x=0, is the solution.
Suppose cotx+cscx=0 .
This states that i(eix+e−ix)eix−e−ix+2ieix−e−ix=0 . the denominators are equal, we need only to equate the sum of the numerators to zero, i.e., i(eix+e−ix)+2i=0 .
Dividing by i , we get eix+2+e−ix=0 .
The left-hand side is nearly thesameastheequationabove(eix/2+e−ix/2)2 , therefore eix/2+e−ix/2=0isverysimilar.
assuming u=eix/2 then u+1u=0 , impling u2+1=0 .
assuming It follows that u=±i . , [math]e^{[/math]
Easy way: There are no solutions. If cotx+cscx=0 , then multiplying both sides by cscx−cotx , we get csc2x−cot2x=0 . But csc2x−cot2x=1 for any x (this is a trig identity):
csc2x−cot2x=1sin2x−cos2xsin2x=1−cos2xsin2x=sin2xsin2x=1
and so we have a contradiction.
Hard way: Suppose cotx+cscx=0 .
This states that i(eix+e−ix)eix−e−ix+2ieix−e−ix=0 . Since the denominators are equal, we need only equate the sum of the numerators to zero, i.e., i(eix+e−ix)+2i=0 .
Dividing by i , we get eix+2+e−ix=0 .
The left-hand side is merely (eix/2+e−ix/2)2 , and therefore eix/2+e−ix/2=0 .
If u=eix/2 then u+1u=0 , and therefore u2+1=0 . It follows that u=±i . Therefore, eix/2=±i , which implies (by taking logarithms) that ix2=iπ2+πik for some k∈Z . From there, we conclude that x=π+2πk for some k∈Z .
All these solutions are actually extraneous, because if x is an integer multiple of π , then sinx=0 so that cotx=cosxsinx and cscx=1sinx are both undefined (division by zero). So there are no solutions.
cot(x) + csc(x) = 0
csc(x) = 1/(sin(x)), cot(x) = cos(x)/sin(x)
because they have the same denominator,
(cos(x) + 1)/sin(x) = 0
since anything multiplied by zero, and vice versa, is equal to zero, we can ignore the denominator. Only problem is, there might be a lost solution by doing that.
ignoring that fact, we get
cos(x) + 1 = 0
cos(x) = -1
cos^-1(cos(x)) = cos^-1(-1)
x = pi
Now if we change the process to be,
cos(x)/sin(x) + 1/sin(x) = 0
cos(x)/sin(x) = -1/sin(x)
multiply sine on both sides,
cos(x) = -1
we end up where we started.
Which means the answer can only be (pi). Including degrees, that would give the answer of 180 degrees. Depends on what you’re looking for though.
Checking the answer on a graph, this is the only correct answer.
Solution: x = Pi
Answer:
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