Simplify $frac {sec^2theta - cos^2theta}{tan^2theta}$











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My question is to simplify:$$frac {sec^2theta - cos^2theta}{tan^2theta}$$





My workings:



$$frac {sec^2theta - cos^2theta}{tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}$$



Then I am unable to continue, can someone please show me some working outs? Thank you.






trigonometry







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  • 1




    $$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
    – Nosrati
    Nov 23 at 7:15












  • $dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
    – Narasimham
    Nov 24 at 15:38

















up vote
0
down vote

favorite













My question is to simplify:$$frac {sec^2theta - cos^2theta}{tan^2theta}$$





My workings:



$$frac {sec^2theta - cos^2theta}{tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}$$



Then I am unable to continue, can someone please show me some working outs? Thank you.






trigonometry







share|cite|improve this question




















  • 1




    $$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
    – Nosrati
    Nov 23 at 7:15












  • $dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
    – Narasimham
    Nov 24 at 15:38















up vote
0
down vote

favorite









up vote
0
down vote

favorite












My question is to simplify:$$frac {sec^2theta - cos^2theta}{tan^2theta}$$





My workings:



$$frac {sec^2theta - cos^2theta}{tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}$$



Then I am unable to continue, can someone please show me some working outs? Thank you.






trigonometry







share|cite|improve this question
















My question is to simplify:$$frac {sec^2theta - cos^2theta}{tan^2theta}$$





My workings:



$$frac {sec^2theta - cos^2theta}{tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}$$



Then I am unable to continue, can someone please show me some working outs? Thank you.






trigonometry




trigonometry






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edited Nov 23 at 7:27









Jean-Claude Arbaut

14.9k63362




14.9k63362










asked Nov 23 at 7:13









Tfue

807




807








  • 1




    $$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
    – Nosrati
    Nov 23 at 7:15












  • $dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
    – Narasimham
    Nov 24 at 15:38
















  • 1




    $$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
    – Nosrati
    Nov 23 at 7:15












  • $dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
    – Narasimham
    Nov 24 at 15:38










1




1




$$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
– Nosrati
Nov 23 at 7:15






$$dfrac{1}{cos^2t}-cos^2t=dfrac{1-cos^4t}{cos^2t}=dfrac{(1-cos^2t)(1+cos^2t)}{cos^2t}=dfrac{(sin^2t)(1+cos^2t)}{cos^2t}=tan^2t(1+cos^2t)$$
– Nosrati
Nov 23 at 7:15














$dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
– Narasimham
Nov 24 at 15:38






$dfrac{1/c^2-c^2}{s^2/c^2}=dfrac{1-c^4}{s^2}=1+c^2$
– Narasimham
Nov 24 at 15:38












2 Answers
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up vote
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accepted










$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$



$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$



From here, use the common denominator $cos^2 theta$.



$$=frac{frac{1}{cos^2 theta}-frac{cos^4 theta}{cos^2 theta}}{tan^2 theta}$$



$$=frac{frac{1-cos^4 theta}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{frac{sin^2 theta}{cos^2 theta}} = frac{{(1-cos^2 theta)(1+cos^2 theta)}}{sin^2 theta}$$



Recall $1-cos^2theta = sin^2 theta$.



$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$






share|cite|improve this answer





















  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Tfue
    Nov 23 at 7:27








  • 1




    You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
    – KM101
    Nov 23 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Tfue
    Nov 23 at 7:30










  • Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
    – KM101
    Nov 23 at 7:33




















up vote
3
down vote













For $cos theta neq 0$ we have



$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$






share|cite|improve this answer





















  • Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
    – Tfue
    Nov 23 at 7:23








  • 2




    @Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
    – gimusi
    Nov 23 at 7:24






  • 1




    You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
    – KM101
    Nov 23 at 7:25








  • 1




    @Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
    – gimusi
    Nov 23 at 7:27











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2 Answers
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active

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2 Answers
2






active

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active

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up vote
3
down vote



accepted










$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$



$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$



From here, use the common denominator $cos^2 theta$.



$$=frac{frac{1}{cos^2 theta}-frac{cos^4 theta}{cos^2 theta}}{tan^2 theta}$$



$$=frac{frac{1-cos^4 theta}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{frac{sin^2 theta}{cos^2 theta}} = frac{{(1-cos^2 theta)(1+cos^2 theta)}}{sin^2 theta}$$



Recall $1-cos^2theta = sin^2 theta$.



$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$






share|cite|improve this answer





















  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Tfue
    Nov 23 at 7:27








  • 1




    You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
    – KM101
    Nov 23 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Tfue
    Nov 23 at 7:30










  • Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
    – KM101
    Nov 23 at 7:33

















up vote
3
down vote



accepted










$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$



$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$



From here, use the common denominator $cos^2 theta$.



$$=frac{frac{1}{cos^2 theta}-frac{cos^4 theta}{cos^2 theta}}{tan^2 theta}$$



$$=frac{frac{1-cos^4 theta}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{frac{sin^2 theta}{cos^2 theta}} = frac{{(1-cos^2 theta)(1+cos^2 theta)}}{sin^2 theta}$$



Recall $1-cos^2theta = sin^2 theta$.



$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$






share|cite|improve this answer





















  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Tfue
    Nov 23 at 7:27








  • 1




    You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
    – KM101
    Nov 23 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Tfue
    Nov 23 at 7:30










  • Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
    – KM101
    Nov 23 at 7:33















up vote
3
down vote



accepted







up vote
3
down vote



accepted






$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$



$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$



From here, use the common denominator $cos^2 theta$.



$$=frac{frac{1}{cos^2 theta}-frac{cos^4 theta}{cos^2 theta}}{tan^2 theta}$$



$$=frac{frac{1-cos^4 theta}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{frac{sin^2 theta}{cos^2 theta}} = frac{{(1-cos^2 theta)(1+cos^2 theta)}}{sin^2 theta}$$



Recall $1-cos^2theta = sin^2 theta$.



$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$






share|cite|improve this answer












$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$



$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$



From here, use the common denominator $cos^2 theta$.



$$=frac{frac{1}{cos^2 theta}-frac{cos^4 theta}{cos^2 theta}}{tan^2 theta}$$



$$=frac{frac{1-cos^4 theta}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{tan^2 theta} = frac{frac{(1-cos^2 theta)(1+cos^2 theta)}{cos^2 theta}}{frac{sin^2 theta}{cos^2 theta}} = frac{{(1-cos^2 theta)(1+cos^2 theta)}}{sin^2 theta}$$



Recall $1-cos^2theta = sin^2 theta$.



$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 7:22









KM101

2,527416




2,527416












  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Tfue
    Nov 23 at 7:27








  • 1




    You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
    – KM101
    Nov 23 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Tfue
    Nov 23 at 7:30










  • Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
    – KM101
    Nov 23 at 7:33




















  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Tfue
    Nov 23 at 7:27








  • 1




    You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
    – KM101
    Nov 23 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Tfue
    Nov 23 at 7:30










  • Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
    – KM101
    Nov 23 at 7:33


















Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
– Tfue
Nov 23 at 7:27






Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
– Tfue
Nov 23 at 7:27






1




1




You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
– KM101
Nov 23 at 7:28






You mean when I factored it to get $(1-cos^2 theta)(1-sin^2 theta)$? If so, that is simply using the difference of squares. It is helpful to remember that $a^2-b^2 = (a+b)(a-b)$.
– KM101
Nov 23 at 7:28














So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
– Tfue
Nov 23 at 7:30




So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
– Tfue
Nov 23 at 7:30












Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
– KM101
Nov 23 at 7:33






Yes to both questions. Of course, as gimusi said, there are many ways (which are essentially equivalent) to simplify trig expressions since there are many definitions and base identities which lead to the same answer. It's good to use whatever way you find simplest.
– KM101
Nov 23 at 7:33












up vote
3
down vote













For $cos theta neq 0$ we have



$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$






share|cite|improve this answer





















  • Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
    – Tfue
    Nov 23 at 7:23








  • 2




    @Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
    – gimusi
    Nov 23 at 7:24






  • 1




    You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
    – KM101
    Nov 23 at 7:25








  • 1




    @Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
    – gimusi
    Nov 23 at 7:27















up vote
3
down vote













For $cos theta neq 0$ we have



$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$






share|cite|improve this answer





















  • Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
    – Tfue
    Nov 23 at 7:23








  • 2




    @Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
    – gimusi
    Nov 23 at 7:24






  • 1




    You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
    – KM101
    Nov 23 at 7:25








  • 1




    @Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
    – gimusi
    Nov 23 at 7:27













up vote
3
down vote










up vote
3
down vote









For $cos theta neq 0$ we have



$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$






share|cite|improve this answer












For $cos theta neq 0$ we have



$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 7:17









gimusi

88.2k74394




88.2k74394












  • Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
    – Tfue
    Nov 23 at 7:23








  • 2




    @Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
    – gimusi
    Nov 23 at 7:24






  • 1




    You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
    – KM101
    Nov 23 at 7:25








  • 1




    @Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
    – gimusi
    Nov 23 at 7:27


















  • Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
    – Tfue
    Nov 23 at 7:23








  • 2




    @Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
    – gimusi
    Nov 23 at 7:24






  • 1




    You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
    – KM101
    Nov 23 at 7:25








  • 1




    @Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
    – gimusi
    Nov 23 at 7:27
















Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
– Tfue
Nov 23 at 7:23






Hey thanks! but why did you times $frac {cos^2theta}{cos^2theta}$ to the equation, how does it works and can you do that? Also can i apply your technique in all cases?
– Tfue
Nov 23 at 7:23






2




2




@Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
– gimusi
Nov 23 at 7:24




@Tfue We can do that since it is equal to 1 and for the definition of the expression we are assuming $cos theta neq 0$.
– gimusi
Nov 23 at 7:24




1




1




You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
– KM101
Nov 23 at 7:25






You can divide/multiply both the numerator and denominator by $cos theta$ to rewrite the expression in a form that can be simplified more easily. (Essentially, you're multiplying/dividing by $1$, so it is equivalent to the original expression.)
– KM101
Nov 23 at 7:25






1




1




@Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
– gimusi
Nov 23 at 7:27




@Tfue Of course we can proceed in many others equivalent ways, for example $$frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac {frac {1}{cos^2theta}- cos^2theta} {frac{sin^2theta}{cos^2theta}}=frac{cos^2theta}{sin^2theta}left(frac {1}{cos^2theta}- cos^2thetaright)=ldots$$
– gimusi
Nov 23 at 7:27


















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