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
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
$endgroup$
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
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
$endgroup$
$begingroup$
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
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
$endgroup$
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
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
edited Feb 3 '14 at 0:58
user127.0.0.1
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
edited Feb 3 '14 at 0:58
user127.0.0.1
5,95162139
edited Feb 3 '14 at 0:58
user127.0.0.1
5,95162139
edited Feb 3 '14 at 0:58
user127.0.0.1
5,95162139
5,95162139
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
asked Feb 3 '14 at 0:43
LearnerLearner
34941022
34941022
$begingroup$
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
8,06121738
$endgroup$
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
6,19611526
$endgroup$
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.
25k22539
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f661439%2fprove-trig-identity-tan-cot-sec-csc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
8,06121738
$endgroup$
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
8,06121738
$endgroup$
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
8,06121738
$endgroup$
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
8,06121738
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,06121738
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,06121738
answered Feb 3 '14 at 0:50
mathematics2x2lifemathematics2x2life
8,06121738
8,06121738
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
6,19611526
$endgroup$
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
6,19611526
$endgroup$
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
6,19611526
$endgroup$
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
6,19611526
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,19611526
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,19611526
answered Aug 11 '14 at 20:51
John JoyJohn Joy
6,19611526
6,19611526
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.
25k22539
$endgroup$
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.
25k22539
$endgroup$
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.
25k22539
$endgroup$
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.
25k22539
edited Dec 19 '18 at 17:24
answered Dec 19 '18 at 17:17
J.G.J.G.
25k22539
answered Dec 19 '18 at 17:17
J.G.J.G.
25k22539
answered Dec 19 '18 at 17:17
J.G.J.G.
25k22539
25k22539
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f661439%2fprove-trig-identity-tan-cot-sec-csc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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