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How does one define restrictions on $x$ or $theta$ when simplifying trigonometric identities?
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When simplifying a trigonometric identity, such as $fracsec^2(x)tan(x)$, how does one find the domain restrictions for these equations? Is it simply all values for which any trig function is either $0$ or undefined? When working through the problem, I found that $xneq tan^-1(0)$, $cos^-1(0)$, $sin^-1(0)$, or $cot^-1(0)$, but isn't it theoretically possible to get any trigonometric function in a denominator, thereby adding another restriction?
trigonometry
edited Jul 31 at 22:26
N. F. Taussig
38k93053
asked Jul 31 at 18:25
Matthew Sylvester
335
add a comment |Â
up vote
1
down vote
favorite
When simplifying a trigonometric identity, such as $fracsec^2(x)tan(x)$, how does one find the domain restrictions for these equations? Is it simply all values for which any trig function is either $0$ or undefined? When working through the problem, I found that $xneq tan^-1(0)$, $cos^-1(0)$, $sin^-1(0)$, or $cot^-1(0)$, but isn't it theoretically possible to get any trigonometric function in a denominator, thereby adding another restriction?
trigonometry
edited Jul 31 at 22:26
N. F. Taussig
38k93053
asked Jul 31 at 18:25
Matthew Sylvester
335
The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When simplifying a trigonometric identity, such as $fracsec^2(x)tan(x)$, how does one find the domain restrictions for these equations? Is it simply all values for which any trig function is either $0$ or undefined? When working through the problem, I found that $xneq tan^-1(0)$, $cos^-1(0)$, $sin^-1(0)$, or $cot^-1(0)$, but isn't it theoretically possible to get any trigonometric function in a denominator, thereby adding another restriction?
trigonometry
edited Jul 31 at 22:26
N. F. Taussig
38k93053
asked Jul 31 at 18:25
Matthew Sylvester
335
When simplifying a trigonometric identity, such as $fracsec^2(x)tan(x)$, how does one find the domain restrictions for these equations? Is it simply all values for which any trig function is either $0$ or undefined? When working through the problem, I found that $xneq tan^-1(0)$, $cos^-1(0)$, $sin^-1(0)$, or $cot^-1(0)$, but isn't it theoretically possible to get any trigonometric function in a denominator, thereby adding another restriction?
trigonometry
edited Jul 31 at 22:26
N. F. Taussig
38k93053
asked Jul 31 at 18:25
Matthew Sylvester
335
edited Jul 31 at 22:26
N. F. Taussig
38k93053
edited Jul 31 at 22:26
N. F. Taussig
38k93053
edited Jul 31 at 22:26
N. F. Taussig
38k93053
38k93053
asked Jul 31 at 18:25
Matthew Sylvester
335
asked Jul 31 at 18:25
Matthew Sylvester
335
asked Jul 31 at 18:25
Matthew Sylvester
335
335
The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49
add a comment |Â
The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49
The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49
The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note. The quantity
$$fracsec^2xtan x$$
is a trigonometric expression, not a trigonometric identity. The equation
$$fracsec^2xtan x = cot x + tan x$$
is a trigonometric identity, meaning that it holds for all values of the variables where both expressions are defined.
Domain of definition of a trigonometric expression
To define the domain of definition of a trigonometric expression, you must make sure that each function in the expression is defined and that you can perform each of the indicated operations.
For the expression
$$fracsec^2xtan x$$
we require that $sec x$ and $tan x$ are defined and that $tan x neq 0$ since $tan x$ is in the denominator.
Since
$$sec x = frac1cos x$$
$sec x$ is only defined when $cos x neq 0$, so we require that
$$x neq fracpi2 + npi, n in mathbbZ$$
Since
$$tan x = fracsin xcos x$$
$tan x$ is only defined when $cos x neq 0$, which leads to the same restriction as above.
Since $tan x = 0 implies x = npi, n in mathbbZ$, we have the additional restriction that $x neq npi, n in mathbbZ$.
Hence, the domain of definition is
$$leftx in mathbbR mid x neq fracpi2 + npi, n in mathbbZright cap leftx in mathbbR mid x neq npi, n in mathbbZright = x in mathbbR mid x neq fracnpi2, n in mathbbZ$$
Domain of definition of a trigonometric identity
To find the domain of definition of a trigonometric identity, we must find the intersection of the domains of each trigonometric expression in the equation are defined.
Example. $dfracsec^2xtan x = cot x + tan x$.
$$fracsec^2xtan x = frac1 + tan^2xtan x = frac1tan x + tan x = cot x + tan x$$
We showed above that the LHS has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Notice that the expression $cot x + tan x$ is only defined when $cot x$ and $tan x$ are both defined.
Since
$$cot x = fraccos xsin x$$
$cot x$ is only defined when $sin x neq 0 implies x neq npi, n in mathbbZ$.
We saw above that $tan x neq 0 implies x neq dfracpi2 + npi, n in mathbbZ$.
Thus, the expression $cot x + tan x$ also has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Therefore, the domain of definition for the identity is
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Example. $dfracsin x1 - cos x = dfrac1 + cos xsin x$.
$$fracsin x1 - cos x = fracsin x1 - cos x cdot frac1 + cos x1 + cos x = fracsin x(1 + cos x)1 - cos^2x = fracsin x(1 + cos x)sin^2x = frac1 + cos xsin x$$
The expression on the left-hand side is defined for all real numbers except
$x = 2npi, n in mathbbZ$. The expression on the right-hand side is defined for all real numbers except $x = npi, n in mathbbZ$. Since both sides of the identity must be defined, the domain of definition of the identity is the intersection of these two domains of definition, which is
$$x in mathbbR mid x neq 2npi, n in mathbbZ cap x in mathbbR mid x neq npi, n in mathbbZ = x in mathbbR mid x neq npi, n in mathbbZ$$
answered Jul 31 at 19:38
N. F. Taussig
38k93053
add a comment |Â
up vote
2
down vote
Yes, it’s possible to add a restriction at any stage if you’re not careful with your working.
As a much simpler example, the function $f(x)=x^2$ obviously has no natural restrictions to its domain. But $g(x)=fracx^2(x-5)x-5$, although you could simplify the fraction making it identical to $f(x)$, would have to be restricted at $x=5$. You can see how we could introduce any number of arbitrary restrictions.
When working with $fracmathrmsec^2xmathrmtanx$, just try to avoid dividing by anything unnecessary. However, even if you introduce arbitrary restrictions along the way, you can check each of them individually at the end.
answered Jul 31 at 18:44
Malkin
822419
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note. The quantity
$$fracsec^2xtan x$$
is a trigonometric expression, not a trigonometric identity. The equation
$$fracsec^2xtan x = cot x + tan x$$
is a trigonometric identity, meaning that it holds for all values of the variables where both expressions are defined.
Domain of definition of a trigonometric expression
To define the domain of definition of a trigonometric expression, you must make sure that each function in the expression is defined and that you can perform each of the indicated operations.
For the expression
$$fracsec^2xtan x$$
we require that $sec x$ and $tan x$ are defined and that $tan x neq 0$ since $tan x$ is in the denominator.
Since
$$sec x = frac1cos x$$
$sec x$ is only defined when $cos x neq 0$, so we require that
$$x neq fracpi2 + npi, n in mathbbZ$$
Since
$$tan x = fracsin xcos x$$
$tan x$ is only defined when $cos x neq 0$, which leads to the same restriction as above.
Since $tan x = 0 implies x = npi, n in mathbbZ$, we have the additional restriction that $x neq npi, n in mathbbZ$.
Hence, the domain of definition is
$$leftx in mathbbR mid x neq fracpi2 + npi, n in mathbbZright cap leftx in mathbbR mid x neq npi, n in mathbbZright = x in mathbbR mid x neq fracnpi2, n in mathbbZ$$
Domain of definition of a trigonometric identity
To find the domain of definition of a trigonometric identity, we must find the intersection of the domains of each trigonometric expression in the equation are defined.
Example. $dfracsec^2xtan x = cot x + tan x$.
$$fracsec^2xtan x = frac1 + tan^2xtan x = frac1tan x + tan x = cot x + tan x$$
We showed above that the LHS has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Notice that the expression $cot x + tan x$ is only defined when $cot x$ and $tan x$ are both defined.
Since
$$cot x = fraccos xsin x$$
$cot x$ is only defined when $sin x neq 0 implies x neq npi, n in mathbbZ$.
We saw above that $tan x neq 0 implies x neq dfracpi2 + npi, n in mathbbZ$.
Thus, the expression $cot x + tan x$ also has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Therefore, the domain of definition for the identity is
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Example. $dfracsin x1 - cos x = dfrac1 + cos xsin x$.
$$fracsin x1 - cos x = fracsin x1 - cos x cdot frac1 + cos x1 + cos x = fracsin x(1 + cos x)1 - cos^2x = fracsin x(1 + cos x)sin^2x = frac1 + cos xsin x$$
The expression on the left-hand side is defined for all real numbers except
$x = 2npi, n in mathbbZ$. The expression on the right-hand side is defined for all real numbers except $x = npi, n in mathbbZ$. Since both sides of the identity must be defined, the domain of definition of the identity is the intersection of these two domains of definition, which is
$$x in mathbbR mid x neq 2npi, n in mathbbZ cap x in mathbbR mid x neq npi, n in mathbbZ = x in mathbbR mid x neq npi, n in mathbbZ$$
answered Jul 31 at 19:38
N. F. Taussig
38k93053
add a comment |Â
up vote
1
down vote
accepted
Note. The quantity
$$fracsec^2xtan x$$
is a trigonometric expression, not a trigonometric identity. The equation
$$fracsec^2xtan x = cot x + tan x$$
is a trigonometric identity, meaning that it holds for all values of the variables where both expressions are defined.
Domain of definition of a trigonometric expression
To define the domain of definition of a trigonometric expression, you must make sure that each function in the expression is defined and that you can perform each of the indicated operations.
For the expression
$$fracsec^2xtan x$$
we require that $sec x$ and $tan x$ are defined and that $tan x neq 0$ since $tan x$ is in the denominator.
Since
$$sec x = frac1cos x$$
$sec x$ is only defined when $cos x neq 0$, so we require that
$$x neq fracpi2 + npi, n in mathbbZ$$
Since
$$tan x = fracsin xcos x$$
$tan x$ is only defined when $cos x neq 0$, which leads to the same restriction as above.
Since $tan x = 0 implies x = npi, n in mathbbZ$, we have the additional restriction that $x neq npi, n in mathbbZ$.
Hence, the domain of definition is
$$leftx in mathbbR mid x neq fracpi2 + npi, n in mathbbZright cap leftx in mathbbR mid x neq npi, n in mathbbZright = x in mathbbR mid x neq fracnpi2, n in mathbbZ$$
Domain of definition of a trigonometric identity
To find the domain of definition of a trigonometric identity, we must find the intersection of the domains of each trigonometric expression in the equation are defined.
Example. $dfracsec^2xtan x = cot x + tan x$.
$$fracsec^2xtan x = frac1 + tan^2xtan x = frac1tan x + tan x = cot x + tan x$$
We showed above that the LHS has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Notice that the expression $cot x + tan x$ is only defined when $cot x$ and $tan x$ are both defined.
Since
$$cot x = fraccos xsin x$$
$cot x$ is only defined when $sin x neq 0 implies x neq npi, n in mathbbZ$.
We saw above that $tan x neq 0 implies x neq dfracpi2 + npi, n in mathbbZ$.
Thus, the expression $cot x + tan x$ also has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Therefore, the domain of definition for the identity is
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Example. $dfracsin x1 - cos x = dfrac1 + cos xsin x$.
$$fracsin x1 - cos x = fracsin x1 - cos x cdot frac1 + cos x1 + cos x = fracsin x(1 + cos x)1 - cos^2x = fracsin x(1 + cos x)sin^2x = frac1 + cos xsin x$$
The expression on the left-hand side is defined for all real numbers except
$x = 2npi, n in mathbbZ$. The expression on the right-hand side is defined for all real numbers except $x = npi, n in mathbbZ$. Since both sides of the identity must be defined, the domain of definition of the identity is the intersection of these two domains of definition, which is
$$x in mathbbR mid x neq 2npi, n in mathbbZ cap x in mathbbR mid x neq npi, n in mathbbZ = x in mathbbR mid x neq npi, n in mathbbZ$$
answered Jul 31 at 19:38
N. F. Taussig
38k93053
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note. The quantity
$$fracsec^2xtan x$$
is a trigonometric expression, not a trigonometric identity. The equation
$$fracsec^2xtan x = cot x + tan x$$
is a trigonometric identity, meaning that it holds for all values of the variables where both expressions are defined.
Domain of definition of a trigonometric expression
To define the domain of definition of a trigonometric expression, you must make sure that each function in the expression is defined and that you can perform each of the indicated operations.
For the expression
$$fracsec^2xtan x$$
we require that $sec x$ and $tan x$ are defined and that $tan x neq 0$ since $tan x$ is in the denominator.
Since
$$sec x = frac1cos x$$
$sec x$ is only defined when $cos x neq 0$, so we require that
$$x neq fracpi2 + npi, n in mathbbZ$$
Since
$$tan x = fracsin xcos x$$
$tan x$ is only defined when $cos x neq 0$, which leads to the same restriction as above.
Since $tan x = 0 implies x = npi, n in mathbbZ$, we have the additional restriction that $x neq npi, n in mathbbZ$.
Hence, the domain of definition is
$$leftx in mathbbR mid x neq fracpi2 + npi, n in mathbbZright cap leftx in mathbbR mid x neq npi, n in mathbbZright = x in mathbbR mid x neq fracnpi2, n in mathbbZ$$
Domain of definition of a trigonometric identity
To find the domain of definition of a trigonometric identity, we must find the intersection of the domains of each trigonometric expression in the equation are defined.
Example. $dfracsec^2xtan x = cot x + tan x$.
$$fracsec^2xtan x = frac1 + tan^2xtan x = frac1tan x + tan x = cot x + tan x$$
We showed above that the LHS has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Notice that the expression $cot x + tan x$ is only defined when $cot x$ and $tan x$ are both defined.
Since
$$cot x = fraccos xsin x$$
$cot x$ is only defined when $sin x neq 0 implies x neq npi, n in mathbbZ$.
We saw above that $tan x neq 0 implies x neq dfracpi2 + npi, n in mathbbZ$.
Thus, the expression $cot x + tan x$ also has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Therefore, the domain of definition for the identity is
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Example. $dfracsin x1 - cos x = dfrac1 + cos xsin x$.
$$fracsin x1 - cos x = fracsin x1 - cos x cdot frac1 + cos x1 + cos x = fracsin x(1 + cos x)1 - cos^2x = fracsin x(1 + cos x)sin^2x = frac1 + cos xsin x$$
The expression on the left-hand side is defined for all real numbers except
$x = 2npi, n in mathbbZ$. The expression on the right-hand side is defined for all real numbers except $x = npi, n in mathbbZ$. Since both sides of the identity must be defined, the domain of definition of the identity is the intersection of these two domains of definition, which is
$$x in mathbbR mid x neq 2npi, n in mathbbZ cap x in mathbbR mid x neq npi, n in mathbbZ = x in mathbbR mid x neq npi, n in mathbbZ$$
answered Jul 31 at 19:38
N. F. Taussig
38k93053
Note. The quantity
$$fracsec^2xtan x$$
is a trigonometric expression, not a trigonometric identity. The equation
$$fracsec^2xtan x = cot x + tan x$$
is a trigonometric identity, meaning that it holds for all values of the variables where both expressions are defined.
Domain of definition of a trigonometric expression
To define the domain of definition of a trigonometric expression, you must make sure that each function in the expression is defined and that you can perform each of the indicated operations.
For the expression
$$fracsec^2xtan x$$
we require that $sec x$ and $tan x$ are defined and that $tan x neq 0$ since $tan x$ is in the denominator.
Since
$$sec x = frac1cos x$$
$sec x$ is only defined when $cos x neq 0$, so we require that
$$x neq fracpi2 + npi, n in mathbbZ$$
Since
$$tan x = fracsin xcos x$$
$tan x$ is only defined when $cos x neq 0$, which leads to the same restriction as above.
Since $tan x = 0 implies x = npi, n in mathbbZ$, we have the additional restriction that $x neq npi, n in mathbbZ$.
Hence, the domain of definition is
$$leftx in mathbbR mid x neq fracpi2 + npi, n in mathbbZright cap leftx in mathbbR mid x neq npi, n in mathbbZright = x in mathbbR mid x neq fracnpi2, n in mathbbZ$$
Domain of definition of a trigonometric identity
To find the domain of definition of a trigonometric identity, we must find the intersection of the domains of each trigonometric expression in the equation are defined.
Example. $dfracsec^2xtan x = cot x + tan x$.
$$fracsec^2xtan x = frac1 + tan^2xtan x = frac1tan x + tan x = cot x + tan x$$
We showed above that the LHS has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Notice that the expression $cot x + tan x$ is only defined when $cot x$ and $tan x$ are both defined.
Since
$$cot x = fraccos xsin x$$
$cot x$ is only defined when $sin x neq 0 implies x neq npi, n in mathbbZ$.
We saw above that $tan x neq 0 implies x neq dfracpi2 + npi, n in mathbbZ$.
Thus, the expression $cot x + tan x$ also has domain of definition
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Therefore, the domain of definition for the identity is
$$leftx in mathbbR mid x neq fracnpi2, n in mathbbZright$$
Example. $dfracsin x1 - cos x = dfrac1 + cos xsin x$.
$$fracsin x1 - cos x = fracsin x1 - cos x cdot frac1 + cos x1 + cos x = fracsin x(1 + cos x)1 - cos^2x = fracsin x(1 + cos x)sin^2x = frac1 + cos xsin x$$
The expression on the left-hand side is defined for all real numbers except
$x = 2npi, n in mathbbZ$. The expression on the right-hand side is defined for all real numbers except $x = npi, n in mathbbZ$. Since both sides of the identity must be defined, the domain of definition of the identity is the intersection of these two domains of definition, which is
$$x in mathbbR mid x neq 2npi, n in mathbbZ cap x in mathbbR mid x neq npi, n in mathbbZ = x in mathbbR mid x neq npi, n in mathbbZ$$
answered Jul 31 at 19:38
N. F. Taussig
38k93053
edited Jul 31 at 22:41
answered Jul 31 at 19:38
N. F. Taussig
38k93053
answered Jul 31 at 19:38
N. F. Taussig
38k93053
answered Jul 31 at 19:38
N. F. Taussig
38k93053
38k93053
add a comment |Â
add a comment |Â
up vote
2
down vote
Yes, it’s possible to add a restriction at any stage if you’re not careful with your working.
As a much simpler example, the function $f(x)=x^2$ obviously has no natural restrictions to its domain. But $g(x)=fracx^2(x-5)x-5$, although you could simplify the fraction making it identical to $f(x)$, would have to be restricted at $x=5$. You can see how we could introduce any number of arbitrary restrictions.
When working with $fracmathrmsec^2xmathrmtanx$, just try to avoid dividing by anything unnecessary. However, even if you introduce arbitrary restrictions along the way, you can check each of them individually at the end.
answered Jul 31 at 18:44
Malkin
822419
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
add a comment |Â
up vote
2
down vote
Yes, it’s possible to add a restriction at any stage if you’re not careful with your working.
As a much simpler example, the function $f(x)=x^2$ obviously has no natural restrictions to its domain. But $g(x)=fracx^2(x-5)x-5$, although you could simplify the fraction making it identical to $f(x)$, would have to be restricted at $x=5$. You can see how we could introduce any number of arbitrary restrictions.
When working with $fracmathrmsec^2xmathrmtanx$, just try to avoid dividing by anything unnecessary. However, even if you introduce arbitrary restrictions along the way, you can check each of them individually at the end.
answered Jul 31 at 18:44
Malkin
822419
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, it’s possible to add a restriction at any stage if you’re not careful with your working.
As a much simpler example, the function $f(x)=x^2$ obviously has no natural restrictions to its domain. But $g(x)=fracx^2(x-5)x-5$, although you could simplify the fraction making it identical to $f(x)$, would have to be restricted at $x=5$. You can see how we could introduce any number of arbitrary restrictions.
When working with $fracmathrmsec^2xmathrmtanx$, just try to avoid dividing by anything unnecessary. However, even if you introduce arbitrary restrictions along the way, you can check each of them individually at the end.
answered Jul 31 at 18:44
Malkin
822419
Yes, it’s possible to add a restriction at any stage if you’re not careful with your working.
As a much simpler example, the function $f(x)=x^2$ obviously has no natural restrictions to its domain. But $g(x)=fracx^2(x-5)x-5$, although you could simplify the fraction making it identical to $f(x)$, would have to be restricted at $x=5$. You can see how we could introduce any number of arbitrary restrictions.
When working with $fracmathrmsec^2xmathrmtanx$, just try to avoid dividing by anything unnecessary. However, even if you introduce arbitrary restrictions along the way, you can check each of them individually at the end.
answered Jul 31 at 18:44
Malkin
822419
answered Jul 31 at 18:44
Malkin
822419
answered Jul 31 at 18:44
Malkin
822419
answered Jul 31 at 18:44
Malkin
822419
822419
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
add a comment |Â
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
What do you mean by "check them off?"
– Matthew Sylvester
Aug 1 at 0:39
add a comment |Â
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The given expression is only defined if both $sec x$ and $tan x$ are defined and $tan x neq 0$. We need the additional restriction on $tan x$ since it is in the denominator.
– N. F. Taussig
Jul 31 at 18:49