Proving $(sec^2x+tan^2x)(csc^2x+cot^2x)=1+2sec^2xcsc^2x$ and $frac{cos x}{1-tan x}+frac{sin x}{1-cot x} = sin...
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Prove the following identities:
$$(sec^2 x + tan^2x)(csc^2 x + cot^2x) = 1+ 2 sec^2x csc^2 x
tag i$$
$$frac{cos x}{1-tan x} + frac{sin x}{1-cot x} = sin x + cos x
tag {ii}$$
For $(mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$
trigonometry
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
$endgroup$
add a comment |
$begingroup$
Prove the following identities:
$$(sec^2 x + tan^2x)(csc^2 x + cot^2x) = 1+ 2 sec^2x csc^2 x
tag i$$
$$frac{cos x}{1-tan x} + frac{sin x}{1-cot x} = sin x + cos x
tag {ii}$$
For $(mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$
trigonometry
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:34
1
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I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
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– Mefitico
Dec 5 '18 at 20:34
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Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
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– N. F. Taussig
Dec 5 '18 at 21:05
add a comment |
$begingroup$
Prove the following identities:
$$(sec^2 x + tan^2x)(csc^2 x + cot^2x) = 1+ 2 sec^2x csc^2 x
tag i$$
$$frac{cos x}{1-tan x} + frac{sin x}{1-cot x} = sin x + cos x
tag {ii}$$
For $(mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$
trigonometry
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
$endgroup$
Prove the following identities:
$$(sec^2 x + tan^2x)(csc^2 x + cot^2x) = 1+ 2 sec^2x csc^2 x
tag i$$
$$frac{cos x}{1-tan x} + frac{sin x}{1-cot x} = sin x + cos x
tag {ii}$$
For $(mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$
trigonometry
trigonometry
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
edited Dec 5 '18 at 23:51
Lorenzo B.
1,8402520
1,8402520
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
asked Dec 5 '18 at 20:27
S. BejtaS. Bejta
61
61
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:34
1
$begingroup$
I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
$endgroup$
– Mefitico
Dec 5 '18 at 20:34
$begingroup$
Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
$endgroup$
– N. F. Taussig
Dec 5 '18 at 21:05
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:34
1
$begingroup$
I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
$endgroup$
– Mefitico
Dec 5 '18 at 20:34
$begingroup$
Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
$endgroup$
– N. F. Taussig
Dec 5 '18 at 21:05
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:34
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:34
1
1
$begingroup$
I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
$endgroup$
– Mefitico
Dec 5 '18 at 20:34
$begingroup$
I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
$endgroup$
– Mefitico
Dec 5 '18 at 20:34
$begingroup$
Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
$endgroup$
– N. F. Taussig
Dec 5 '18 at 21:05
$begingroup$
Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
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– N. F. Taussig
Dec 5 '18 at 21:05
add a comment |
3 Answers
3
active
oldest
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(i)
$$(sec^2x + tan^2x)(csc^2x + cot^2x)$$
=>
$$(sec^2xcsc^2x + tan^2xcsc^2x + sec^2xcot^2x + tan^2xcot^2x)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x}{cos^2xsin^2x} + frac{cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2x} + frac{1}{sin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x + cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ sec^2xcsc^2x + 1)$$
=>
$$(1 + 2sec^2xcsc^2x)$$
ii) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
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$begingroup$
That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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Thanks! fixed the issues...
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– Gopal Anantharaman
Dec 5 '18 at 21:56
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Well done I lost in that! I'll take a look to your work, Bye
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– gimusi
Dec 5 '18 at 22:02
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It looks good now.
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– Toby Bartels
Dec 6 '18 at 22:00
add a comment |
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$$(sec^2x+tan^2x)(csc^2x+cot^2x)$$
$$=(2sec^2x-1)(2csc^2x-1)$$
$$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$
Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$
The second one has been solved by Taussig
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
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Your method for this proof is definitely an improvement over mine.
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– N. F. Taussig
Dec 8 '18 at 14:53
add a comment |
$begingroup$
(i)
begin{align*}
(sec^2x + tan^2x)(csc^2x + cot^2x) & = (1 + tan^2x + tan^2x)(1 + cot^2x + cot^2x)\
& = (1 + 2tan^2x)(1 + 2cot^2x)\
& = 1 + 2cot^2x + 2tan^2x + 4\
& = 5 + 2(csc^2x - 1) + 2(sec^2x - 1)\
& = 5 + 2csc^2x - 2 + 2sec^2x - 2\
& = 1 + 2csc^2x + 2sec^2x\
& = 1 + 2left(frac{1}{sin^2x} + frac{1}{cos^2x}right)\
& = 1 + 2left(frac{cos^2x + sin^2x}{sin^2xcos^2x}right)\
& = 1 + frac{2}{cos^2xsin^2x}\
& = 1 + 2sec^2xcsc^2x
end{align*}
(ii)
begin{align*}
frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x} & = frac{cos x}{1 - tan x} cdot frac{cos x}{cos x} + frac{sin x}{1 - cot x} cdot frac{sin x}{sin x}\
& = frac{cos^2x}{cos x - sin x} + frac{sin^2x}{sin x - cos x}\
& = frac{cos^2x}{cos x - sin x} - frac{sin^2x}{cos x - sin x}\
& = frac{cos^2x - sin^2x}{cos x - sin x}\
& = frac{(cos x + sin x)(cos x - sin x)}{cos x - sin x}\
& = cos x + sin x
end{align*}
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
(i)
$$(sec^2x + tan^2x)(csc^2x + cot^2x)$$
=>
$$(sec^2xcsc^2x + tan^2xcsc^2x + sec^2xcot^2x + tan^2xcot^2x)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x}{cos^2xsin^2x} + frac{cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2x} + frac{1}{sin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x + cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ sec^2xcsc^2x + 1)$$
=>
$$(1 + 2sec^2xcsc^2x)$$
ii) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
$endgroup$
$begingroup$
That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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Thanks! fixed the issues...
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– Gopal Anantharaman
Dec 5 '18 at 21:56
$begingroup$
Well done I lost in that! I'll take a look to your work, Bye
$endgroup$
– gimusi
Dec 5 '18 at 22:02
$begingroup$
It looks good now.
$endgroup$
– Toby Bartels
Dec 6 '18 at 22:00
add a comment |
$begingroup$
(i)
$$(sec^2x + tan^2x)(csc^2x + cot^2x)$$
=>
$$(sec^2xcsc^2x + tan^2xcsc^2x + sec^2xcot^2x + tan^2xcot^2x)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x}{cos^2xsin^2x} + frac{cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2x} + frac{1}{sin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x + cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ sec^2xcsc^2x + 1)$$
=>
$$(1 + 2sec^2xcsc^2x)$$
ii) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
$endgroup$
$begingroup$
That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
$begingroup$
Thanks! fixed the issues...
$endgroup$
– Gopal Anantharaman
Dec 5 '18 at 21:56
$begingroup$
Well done I lost in that! I'll take a look to your work, Bye
$endgroup$
– gimusi
Dec 5 '18 at 22:02
$begingroup$
It looks good now.
$endgroup$
– Toby Bartels
Dec 6 '18 at 22:00
add a comment |
$begingroup$
(i)
$$(sec^2x + tan^2x)(csc^2x + cot^2x)$$
=>
$$(sec^2xcsc^2x + tan^2xcsc^2x + sec^2xcot^2x + tan^2xcot^2x)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x}{cos^2xsin^2x} + frac{cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2x} + frac{1}{sin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x + cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ sec^2xcsc^2x + 1)$$
=>
$$(1 + 2sec^2xcsc^2x)$$
ii) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
$endgroup$
(i)
$$(sec^2x + tan^2x)(csc^2x + cot^2x)$$
=>
$$(sec^2xcsc^2x + tan^2xcsc^2x + sec^2xcot^2x + tan^2xcot^2x)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x}{cos^2xsin^2x} + frac{cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2x} + frac{1}{sin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{sin^2x + cos^2x}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ frac{1}{cos^2xsin^2x} + 1)$$
=>
$$(sec^2xcsc^2x+ sec^2xcsc^2x + 1)$$
=>
$$(1 + 2sec^2xcsc^2x)$$
ii) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
edited Dec 5 '18 at 21:55
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
answered Dec 5 '18 at 21:21
Gopal AnantharamanGopal Anantharaman
676
676
$begingroup$
That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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Thanks! fixed the issues...
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– Gopal Anantharaman
Dec 5 '18 at 21:56
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Well done I lost in that! I'll take a look to your work, Bye
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– gimusi
Dec 5 '18 at 22:02
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It looks good now.
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– Toby Bartels
Dec 6 '18 at 22:00
add a comment |
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That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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Thanks! fixed the issues...
$endgroup$
– Gopal Anantharaman
Dec 5 '18 at 21:56
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Well done I lost in that! I'll take a look to your work, Bye
$endgroup$
– gimusi
Dec 5 '18 at 22:02
$begingroup$
It looks good now.
$endgroup$
– Toby Bartels
Dec 6 '18 at 22:00
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That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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That's not the desired answer for (ii). Your only mistake that I can see is in going to the line with a fraction inside a fraction; in the numerator there, you don't have a common denominator. That is, the numerator should be $frac{cos xsin x-cos^2x}{sin x}+frac{sin xcos x-sin^2x}{cos x}$, rather than $frac{cos xsin x-cos^2x}{sin xcos x}+frac{sin xcos x-sin^2x}{sin xcos x}=frac{cos xsin x-cos^2x+sin xcos x-sin^2x}{sin xcos x}$ as you have it.
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– Toby Bartels
Dec 5 '18 at 21:34
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Thanks! fixed the issues...
$endgroup$
– Gopal Anantharaman
Dec 5 '18 at 21:56
$begingroup$
Thanks! fixed the issues...
$endgroup$
– Gopal Anantharaman
Dec 5 '18 at 21:56
$begingroup$
Well done I lost in that! I'll take a look to your work, Bye
$endgroup$
– gimusi
Dec 5 '18 at 22:02
$begingroup$
Well done I lost in that! I'll take a look to your work, Bye
$endgroup$
– gimusi
Dec 5 '18 at 22:02
$begingroup$
It looks good now.
$endgroup$
– Toby Bartels
Dec 6 '18 at 22:00
$begingroup$
It looks good now.
$endgroup$
– Toby Bartels
Dec 6 '18 at 22:00
add a comment |
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$$(sec^2x+tan^2x)(csc^2x+cot^2x)$$
$$=(2sec^2x-1)(2csc^2x-1)$$
$$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$
Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$
The second one has been solved by Taussig
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
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Your method for this proof is definitely an improvement over mine.
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– N. F. Taussig
Dec 8 '18 at 14:53
add a comment |
$begingroup$
$$(sec^2x+tan^2x)(csc^2x+cot^2x)$$
$$=(2sec^2x-1)(2csc^2x-1)$$
$$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$
Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$
The second one has been solved by Taussig
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
Your method for this proof is definitely an improvement over mine.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 14:53
add a comment |
$begingroup$
$$(sec^2x+tan^2x)(csc^2x+cot^2x)$$
$$=(2sec^2x-1)(2csc^2x-1)$$
$$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$
Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$
The second one has been solved by Taussig
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$$(sec^2x+tan^2x)(csc^2x+cot^2x)$$
$$=(2sec^2x-1)(2csc^2x-1)$$
$$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$
Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$
The second one has been solved by Taussig
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 6 '18 at 7:41
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
$begingroup$
Your method for this proof is definitely an improvement over mine.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 14:53
add a comment |
$begingroup$
Your method for this proof is definitely an improvement over mine.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 14:53
$begingroup$
Your method for this proof is definitely an improvement over mine.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 14:53
$begingroup$
Your method for this proof is definitely an improvement over mine.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 14:53
add a comment |
$begingroup$
(i)
begin{align*}
(sec^2x + tan^2x)(csc^2x + cot^2x) & = (1 + tan^2x + tan^2x)(1 + cot^2x + cot^2x)\
& = (1 + 2tan^2x)(1 + 2cot^2x)\
& = 1 + 2cot^2x + 2tan^2x + 4\
& = 5 + 2(csc^2x - 1) + 2(sec^2x - 1)\
& = 5 + 2csc^2x - 2 + 2sec^2x - 2\
& = 1 + 2csc^2x + 2sec^2x\
& = 1 + 2left(frac{1}{sin^2x} + frac{1}{cos^2x}right)\
& = 1 + 2left(frac{cos^2x + sin^2x}{sin^2xcos^2x}right)\
& = 1 + frac{2}{cos^2xsin^2x}\
& = 1 + 2sec^2xcsc^2x
end{align*}
(ii)
begin{align*}
frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x} & = frac{cos x}{1 - tan x} cdot frac{cos x}{cos x} + frac{sin x}{1 - cot x} cdot frac{sin x}{sin x}\
& = frac{cos^2x}{cos x - sin x} + frac{sin^2x}{sin x - cos x}\
& = frac{cos^2x}{cos x - sin x} - frac{sin^2x}{cos x - sin x}\
& = frac{cos^2x - sin^2x}{cos x - sin x}\
& = frac{(cos x + sin x)(cos x - sin x)}{cos x - sin x}\
& = cos x + sin x
end{align*}
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
add a comment |
$begingroup$
(i)
begin{align*}
(sec^2x + tan^2x)(csc^2x + cot^2x) & = (1 + tan^2x + tan^2x)(1 + cot^2x + cot^2x)\
& = (1 + 2tan^2x)(1 + 2cot^2x)\
& = 1 + 2cot^2x + 2tan^2x + 4\
& = 5 + 2(csc^2x - 1) + 2(sec^2x - 1)\
& = 5 + 2csc^2x - 2 + 2sec^2x - 2\
& = 1 + 2csc^2x + 2sec^2x\
& = 1 + 2left(frac{1}{sin^2x} + frac{1}{cos^2x}right)\
& = 1 + 2left(frac{cos^2x + sin^2x}{sin^2xcos^2x}right)\
& = 1 + frac{2}{cos^2xsin^2x}\
& = 1 + 2sec^2xcsc^2x
end{align*}
(ii)
begin{align*}
frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x} & = frac{cos x}{1 - tan x} cdot frac{cos x}{cos x} + frac{sin x}{1 - cot x} cdot frac{sin x}{sin x}\
& = frac{cos^2x}{cos x - sin x} + frac{sin^2x}{sin x - cos x}\
& = frac{cos^2x}{cos x - sin x} - frac{sin^2x}{cos x - sin x}\
& = frac{cos^2x - sin^2x}{cos x - sin x}\
& = frac{(cos x + sin x)(cos x - sin x)}{cos x - sin x}\
& = cos x + sin x
end{align*}
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
add a comment |
$begingroup$
(i)
begin{align*}
(sec^2x + tan^2x)(csc^2x + cot^2x) & = (1 + tan^2x + tan^2x)(1 + cot^2x + cot^2x)\
& = (1 + 2tan^2x)(1 + 2cot^2x)\
& = 1 + 2cot^2x + 2tan^2x + 4\
& = 5 + 2(csc^2x - 1) + 2(sec^2x - 1)\
& = 5 + 2csc^2x - 2 + 2sec^2x - 2\
& = 1 + 2csc^2x + 2sec^2x\
& = 1 + 2left(frac{1}{sin^2x} + frac{1}{cos^2x}right)\
& = 1 + 2left(frac{cos^2x + sin^2x}{sin^2xcos^2x}right)\
& = 1 + frac{2}{cos^2xsin^2x}\
& = 1 + 2sec^2xcsc^2x
end{align*}
(ii)
begin{align*}
frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x} & = frac{cos x}{1 - tan x} cdot frac{cos x}{cos x} + frac{sin x}{1 - cot x} cdot frac{sin x}{sin x}\
& = frac{cos^2x}{cos x - sin x} + frac{sin^2x}{sin x - cos x}\
& = frac{cos^2x}{cos x - sin x} - frac{sin^2x}{cos x - sin x}\
& = frac{cos^2x - sin^2x}{cos x - sin x}\
& = frac{(cos x + sin x)(cos x - sin x)}{cos x - sin x}\
& = cos x + sin x
end{align*}
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
(i)
begin{align*}
(sec^2x + tan^2x)(csc^2x + cot^2x) & = (1 + tan^2x + tan^2x)(1 + cot^2x + cot^2x)\
& = (1 + 2tan^2x)(1 + 2cot^2x)\
& = 1 + 2cot^2x + 2tan^2x + 4\
& = 5 + 2(csc^2x - 1) + 2(sec^2x - 1)\
& = 5 + 2csc^2x - 2 + 2sec^2x - 2\
& = 1 + 2csc^2x + 2sec^2x\
& = 1 + 2left(frac{1}{sin^2x} + frac{1}{cos^2x}right)\
& = 1 + 2left(frac{cos^2x + sin^2x}{sin^2xcos^2x}right)\
& = 1 + frac{2}{cos^2xsin^2x}\
& = 1 + 2sec^2xcsc^2x
end{align*}
(ii)
begin{align*}
frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x} & = frac{cos x}{1 - tan x} cdot frac{cos x}{cos x} + frac{sin x}{1 - cot x} cdot frac{sin x}{sin x}\
& = frac{cos^2x}{cos x - sin x} + frac{sin^2x}{sin x - cos x}\
& = frac{cos^2x}{cos x - sin x} - frac{sin^2x}{cos x - sin x}\
& = frac{cos^2x - sin^2x}{cos x - sin x}\
& = frac{(cos x + sin x)(cos x - sin x)}{cos x - sin x}\
& = cos x + sin x
end{align*}
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
answered Dec 5 '18 at 22:25
N. F. TaussigN. F. Taussig
44k93356
44k93356
add a comment |
add a comment |
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Please see math.meta.stackexchange.com/questions/5020
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– Lord Shark the Unknown
Dec 5 '18 at 20:34
1
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I've retyped your equations using mathJax. Notice you can further edit your question, as some parts of the equations were not clear, please do so if something is not as you originally intended.
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– Mefitico
Dec 5 '18 at 20:34
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Welcome to MathSE. Please edit the question to show your work. That helps us identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. You should not have the $sec^2x$ term in your answer to the first question.
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– N. F. Taussig
Dec 5 '18 at 21:05