Mixing
Is $ tan(arctan(frac{1}{x})) $ equal to infinity?
$begingroup$
I am solving the equation below:
$$
S = pi (4 - (p . tan(arctan(frac{1}{p}))^2)
$$
My solution for solving $ tan(arctan(frac{1}{p})) $ is:
$$
tan(arctan(frac{1}{p})) =
tan(frac{pi}{2} - arctan(p)) =
frac{tan(frac{pi}{2}) - tan(arctan(p))}{ 1 + tan(frac{pi}{2})
.tan(arctan(p))} =
$$
$$
frac{∞ - p}{1 + ∞ . p} =
frac{∞}{∞}
$$
I know it is wrong! Please determine my mistake(s) and show me how to solve such equations. thanks!
trigonometry inverse-function
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
$endgroup$
add a comment |
$begingroup$
I am solving the equation below:
$$
S = pi (4 - (p . tan(arctan(frac{1}{p}))^2)
$$
My solution for solving $ tan(arctan(frac{1}{p})) $ is:
$$
tan(arctan(frac{1}{p})) =
tan(frac{pi}{2} - arctan(p)) =
frac{tan(frac{pi}{2}) - tan(arctan(p))}{ 1 + tan(frac{pi}{2})
.tan(arctan(p))} =
$$
$$
frac{∞ - p}{1 + ∞ . p} =
frac{∞}{∞}
$$
I know it is wrong! Please determine my mistake(s) and show me how to solve such equations. thanks!
trigonometry inverse-function
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
$endgroup$
4
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28
add a comment |
$begingroup$
I am solving the equation below:
$$
S = pi (4 - (p . tan(arctan(frac{1}{p}))^2)
$$
My solution for solving $ tan(arctan(frac{1}{p})) $ is:
$$
tan(arctan(frac{1}{p})) =
tan(frac{pi}{2} - arctan(p)) =
frac{tan(frac{pi}{2}) - tan(arctan(p))}{ 1 + tan(frac{pi}{2})
.tan(arctan(p))} =
$$
$$
frac{∞ - p}{1 + ∞ . p} =
frac{∞}{∞}
$$
I know it is wrong! Please determine my mistake(s) and show me how to solve such equations. thanks!
trigonometry inverse-function
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
$endgroup$
I am solving the equation below:
$$
S = pi (4 - (p . tan(arctan(frac{1}{p}))^2)
$$
My solution for solving $ tan(arctan(frac{1}{p})) $ is:
$$
tan(arctan(frac{1}{p})) =
tan(frac{pi}{2} - arctan(p)) =
frac{tan(frac{pi}{2}) - tan(arctan(p))}{ 1 + tan(frac{pi}{2})
.tan(arctan(p))} =
$$
$$
frac{∞ - p}{1 + ∞ . p} =
frac{∞}{∞}
$$
I know it is wrong! Please determine my mistake(s) and show me how to solve such equations. thanks!
trigonometry inverse-function
trigonometry inverse-function
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
asked Jan 4 at 18:21
Amir ForsatiAmir Forsati
1086
1086
4
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28
add a comment |
4
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28
4
4
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not a trig identity wizard but I can tell you that $tan(arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $tan(arctan(frac{1}{p}))$ as that is just $frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
$endgroup$
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I'm not a trig identity wizard but I can tell you that $tan(arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $tan(arctan(frac{1}{p}))$ as that is just $frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
$endgroup$
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
add a comment |
$begingroup$
I'm not a trig identity wizard but I can tell you that $tan(arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $tan(arctan(frac{1}{p}))$ as that is just $frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
$endgroup$
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
add a comment |
$begingroup$
I'm not a trig identity wizard but I can tell you that $tan(arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $tan(arctan(frac{1}{p}))$ as that is just $frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
$endgroup$
I'm not a trig identity wizard but I can tell you that $tan(arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $tan(arctan(frac{1}{p}))$ as that is just $frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
answered Jan 4 at 18:29
Keefer RowanKeefer Rowan
1427
1427
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
add a comment |
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
$begingroup$
Thanks for your answer, I got it. but why is my method wrong?
$endgroup$
– Amir Forsati
Jan 4 at 18:48
add a comment |
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4
$begingroup$
The range of the arctangent is $[-pi/2,pi/2]$. On the domain $(-pi/2,pi/2)$, the range of the tangent is $(-infty,infty)$. Hence, $tan(arctan(x))=x$ for all $xin mathbb{R}$. Is your expression $tan left( arctanleft(frac1pright)times arctanleft(frac1pright)right)?$ Or is your expression $tan left( arctanleft(frac1pright)right)times tan left( arctanleft(frac1pright)right)$
$endgroup$
– Mark Viola
Jan 4 at 18:28