Prove trig identity: $tan + cot = sec csc$
$begingroup$
I appreciate the help.
My attempt:
$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$
trigonometry solution-verification
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
$endgroup$
add a comment |
$begingroup$
I appreciate the help.
My attempt:
$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$
trigonometry solution-verification
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
$endgroup$
$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
1
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49
add a comment |
$begingroup$
I appreciate the help.
My attempt:
$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$
trigonometry solution-verification
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
$endgroup$
I appreciate the help.
My attempt:
$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$
trigonometry solution-verification
trigonometry solution-verification
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
edited Feb 3 '14 at 0:58
user127.0.0.1
5,99162139
5,99162139
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
asked Feb 3 '14 at 0:43
LearnerLearner
35941022
35941022
$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
1
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49
add a comment |
$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
1
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49
$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
1
1
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
$endgroup$
add a comment |
$begingroup$
Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.
Note that
$$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
that
$$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
and that
$$sec^2theta=tan^2+1$$
then
$$begin{array}{lll}
tantheta+cottheta&=&1cdot(tantheta+cottheta)\
&=&(cotthetatantheta)(tantheta+cottheta)\
&=&(cottheta)(tantheta(tantheta+cottheta))\
&=&frac{csctheta}{sectheta}(tan^2theta+1)\
&=&frac{csctheta}{sectheta}sec^2theta\
&=§hetacsctheta
end{array}$$
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
$endgroup$
add a comment |
$begingroup$
An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
$endgroup$
add a comment |
$begingroup$
That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
$endgroup$
add a comment |
$begingroup$
That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
$endgroup$
That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,10621739
8,10621739
add a comment |
add a comment |
$begingroup$
Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.
Note that
$$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
that
$$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
and that
$$sec^2theta=tan^2+1$$
then
$$begin{array}{lll}
tantheta+cottheta&=&1cdot(tantheta+cottheta)\
&=&(cotthetatantheta)(tantheta+cottheta)\
&=&(cottheta)(tantheta(tantheta+cottheta))\
&=&frac{csctheta}{sectheta}(tan^2theta+1)\
&=&frac{csctheta}{sectheta}sec^2theta\
&=§hetacsctheta
end{array}$$
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
$endgroup$
add a comment |
$begingroup$
Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.
Note that
$$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
that
$$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
and that
$$sec^2theta=tan^2+1$$
then
$$begin{array}{lll}
tantheta+cottheta&=&1cdot(tantheta+cottheta)\
&=&(cotthetatantheta)(tantheta+cottheta)\
&=&(cottheta)(tantheta(tantheta+cottheta))\
&=&frac{csctheta}{sectheta}(tan^2theta+1)\
&=&frac{csctheta}{sectheta}sec^2theta\
&=§hetacsctheta
end{array}$$
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
$endgroup$
add a comment |
$begingroup$
Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.
Note that
$$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
that
$$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
and that
$$sec^2theta=tan^2+1$$
then
$$begin{array}{lll}
tantheta+cottheta&=&1cdot(tantheta+cottheta)\
&=&(cotthetatantheta)(tantheta+cottheta)\
&=&(cottheta)(tantheta(tantheta+cottheta))\
&=&frac{csctheta}{sectheta}(tan^2theta+1)\
&=&frac{csctheta}{sectheta}sec^2theta\
&=§hetacsctheta
end{array}$$
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
$endgroup$
Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.
Note that
$$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
that
$$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
and that
$$sec^2theta=tan^2+1$$
then
$$begin{array}{lll}
tantheta+cottheta&=&1cdot(tantheta+cottheta)\
&=&(cotthetatantheta)(tantheta+cottheta)\
&=&(cottheta)(tantheta(tantheta+cottheta))\
&=&frac{csctheta}{sectheta}(tan^2theta+1)\
&=&frac{csctheta}{sectheta}sec^2theta\
&=§hetacsctheta
end{array}$$
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,29911827
6,29911827
add a comment |
add a comment |
$begingroup$
An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
$endgroup$
add a comment |
$begingroup$
An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
$endgroup$
add a comment |
$begingroup$
An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
$endgroup$
An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
edited Dec 19 '18 at 17:24
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
answered Dec 19 '18 at 17:17
J.G.J.G.
32.6k23250
32.6k23250
add a comment |
add a comment |
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$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48
1
$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49