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 $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$










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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




    $begingroup$
    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.
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    – 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.
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    – N. F. Taussig
    Dec 5 '18 at 21:05


















1












$begingroup$



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 $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$










share|cite|improve this question











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  • $begingroup$
    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
















1












1








1





$begingroup$



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 $$1+ sec^2x + frac{2}{cos^2x sin^2x}$$










share|cite|improve this question











$endgroup$





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






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edited Dec 5 '18 at 23:51









Lorenzo B.

1,8402520




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asked Dec 5 '18 at 20:27









S. BejtaS. Bejta

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  • $begingroup$
    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






  • 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.
$endgroup$
– N. F. Taussig
Dec 5 '18 at 21:05












3 Answers
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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) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
=>
$$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
=>
$$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
=>
$$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
=>
$$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
=>
$$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
=>
$$sin x + cos x$$






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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.
    $endgroup$
    – 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



















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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






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    Your method for this proof is definitely an improvement over mine.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 14:53



















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(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*}






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    3 Answers
    3






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    3 Answers
    3






    active

    oldest

    votes









    active

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    active

    oldest

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    2












    $begingroup$

    (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) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
    =>
    $$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
    =>
    $$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
    =>
    $$sin x + cos x$$






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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.
      $endgroup$
      – 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
















    2












    $begingroup$

    (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) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
    =>
    $$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
    =>
    $$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
    =>
    $$sin x + cos x$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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.
      $endgroup$
      – 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














    2












    2








    2





    $begingroup$

    (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) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
    =>
    $$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
    =>
    $$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
    =>
    $$sin x + cos x$$






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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) $$frac{cos x}{1 - tan x} + frac{sin x}{1 - cot x}$$
    =>
    $$frac{cos x(1 - cot x) + sin x(1 - tan x)}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{frac{cos^2x sin x - cos xcos^2x + sin^2 xcos x - sin xsin^2x}{sin x cos x}}{(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{cos^2 x(sin x - cos x) + sin^2 x(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)(cos x - sin x)}{cos xsin x(1 - tan x)(1 - cot x)}$$
    =>
    $$frac{(sin^2 x - cos^2x)(cos x - sin x)}{(cos x - sin x)(sin x - cos x)}$$
    =>
    $$frac{(sin^2 x - cos^2 x)}{(sin x - cos x)}$$
    =>
    $$frac{(sin x + cos x)(sin x - cos x)}{(sin x - cos x)}$$
    =>
    $$sin x + cos x$$







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    edited Dec 5 '18 at 21:55

























    answered Dec 5 '18 at 21:21









    Gopal AnantharamanGopal Anantharaman

    626




    626












    • $begingroup$
      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.
      $endgroup$
      – 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


















    • $begingroup$
      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.
      $endgroup$
      – 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
















    $begingroup$
    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.
    $endgroup$
    – Toby Bartels
    Dec 5 '18 at 21:34






    $begingroup$
    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.
    $endgroup$
    – Toby Bartels
    Dec 5 '18 at 21:34














    $begingroup$
    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











    1












    $begingroup$

    $$(sec^2x+tan^2x)(csc^2x+cot^2x)$$



    $$=(2sec^2x-1)(2csc^2x-1)$$



    $$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$



    Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$



    The second one has been solved by Taussig






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your method for this proof is definitely an improvement over mine.
      $endgroup$
      – N. F. Taussig
      Dec 8 '18 at 14:53
















    1












    $begingroup$

    $$(sec^2x+tan^2x)(csc^2x+cot^2x)$$



    $$=(2sec^2x-1)(2csc^2x-1)$$



    $$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$



    Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$



    The second one has been solved by Taussig






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your method for this proof is definitely an improvement over mine.
      $endgroup$
      – N. F. Taussig
      Dec 8 '18 at 14:53














    1












    1








    1





    $begingroup$

    $$(sec^2x+tan^2x)(csc^2x+cot^2x)$$



    $$=(2sec^2x-1)(2csc^2x-1)$$



    $$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$



    Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$



    The second one has been solved by Taussig






    share|cite|improve this answer









    $endgroup$



    $$(sec^2x+tan^2x)(csc^2x+cot^2x)$$



    $$=(2sec^2x-1)(2csc^2x-1)$$



    $$=4sec^2xcsc^2x-2(sec^2x+csc^2x)+1$$



    Now use $sec^2x+csc^2x=cdots=sec^2xcsc^2x$



    The second one has been solved by Taussig







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 6 '18 at 7:41









    lab bhattacharjeelab bhattacharjee

    224k15156274




    224k15156274












    • $begingroup$
      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




    $begingroup$
    Your method for this proof is definitely an improvement over mine.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 14:53











    0












    $begingroup$

    (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*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      (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*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        (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*}






        share|cite|improve this answer









        $endgroup$



        (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*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 22:25









        N. F. TaussigN. F. Taussig

        44k93355




        44k93355






























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