Josh's Claim
Josh claims that tan θ is undefined for any θ such that sin θ = 1 − a 2 + b 2 b 2 , where b = 0 .
Is Josh correct?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
When a =0 then Here no information about 'a'
Log in to reply
If a = 0 , then sin θ is undefined, and hence tan θ is also undefined.
Here b=0 if a=0 then 0/0 indertminate form will be occur then how can you evaluate 1 -0/0
Log in to reply
That's exactly my point. If a = 0 , then sin θ is undefined, and hence tan θ is also undefined, as stated in the question.
sin@ define everywhere in R
Log in to reply
Yes it is. But consider the value of sin θ where θ → ∞ or θ → − ∞ . You cannot determine such a value. It is simply between − 1 and 1 , inclusive. Therefore, it is not a defined value. Choosing a and b such that both are equal to zero will leave θ in an indeterminate form and thus, the value of sin θ will be undefined.
1/0 is undefined or Infinite But 0/0 is indeterminate only but for 1/0 we can say it will be infinite.
Log in to reply
Division by zero is indeed undefined but I'm afraid I'm not seeing a point here. If the "it" you are referring to is sin θ , then no it won't be infinite since this function is bounded between − 1 and 1 , inclusive ∀ θ . If by "it," you mean 1/0, then it is undefined (infinite), but this brings us right back to the point of the question.
We have that sin θ = 1 − a 2 + b 2 b 2 , and b = 0 . Therefore, sin θ is simply 1 = ± 1 .
When sin θ = 1 or − 1 ⟹ cos θ = 0 , and tan θ is undefined since tan θ = cos θ sin θ .