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












0















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 '18 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 '18 at 15:38


















0















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 '18 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 '18 at 15:38
















0












0








0








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 '18 at 7:27









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Nov 23 '18 at 7:13









Fred WeasleyFred Weasley

1429




1429








  • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 15:38












2 Answers
2






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3














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



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



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 $1-cos^4theta$
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:33





















3














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?
    – Fred Weasley
    Nov 23 '18 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 '18 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 '18 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 '18 at 7:27











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2 Answers
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2 Answers
2






active

oldest

votes









active

oldest

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active

oldest

votes









3














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



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



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 $1-cos^4theta$
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:33


















3














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



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



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 $1-cos^4theta$
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:33
















3












3








3






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



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



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 '18 at 7:22









KM101KM101

5,7411423




5,7411423












  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:33




















  • Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:28












  • So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
    – Fred Weasley
    Nov 23 '18 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 '18 at 7:33


















Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
– Fred Weasley
Nov 23 '18 at 7:27






Wow extremely insightful solution! But i have 1 question, what is the technique when you expand the $1-cos^4theta$
– Fred Weasley
Nov 23 '18 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 '18 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 '18 at 7:28














So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
– Fred Weasley
Nov 23 '18 at 7:30




So i view $b$ as $cos^2theta$ in this case? And does this apply in every cases?
– Fred Weasley
Nov 23 '18 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 '18 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 '18 at 7:33













3














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?
    – Fred Weasley
    Nov 23 '18 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 '18 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 '18 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 '18 at 7:27
















3














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?
    – Fred Weasley
    Nov 23 '18 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 '18 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 '18 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 '18 at 7:27














3












3








3






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 '18 at 7:17









gimusigimusi

1




1












  • 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?
    – Fred Weasley
    Nov 23 '18 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 '18 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 '18 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 '18 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?
    – Fred Weasley
    Nov 23 '18 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 '18 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 '18 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 '18 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?
– Fred Weasley
Nov 23 '18 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?
– Fred Weasley
Nov 23 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 7:27


















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