Simplify $frac {sec^2theta - cos^2theta}{tan^2theta}$
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
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
1429
add a comment |
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
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
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
add a comment |
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
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
1429
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
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
1429
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
1429
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
edited Nov 23 '18 at 7:27
Jean-Claude Arbaut
14.7k63464
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
asked Nov 23 '18 at 7:13
Fred WeasleyFred Weasley
1429
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$
$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$
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 $1-cos^2theta = sin^2 theta$.
$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$
answered Nov 23 '18 at 7:22
KM101KM101
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
add a comment |
For $cos theta neq 0$ we have
$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$
answered Nov 23 '18 at 7:17
gimusigimusi
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
add a comment |
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2 Answers
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2 Answers
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$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$
$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$
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 $1-cos^2theta = sin^2 theta$.
$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$
answered Nov 23 '18 at 7:22
KM101KM101
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
add a comment |
$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$
$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$
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 $1-cos^2theta = sin^2 theta$.
$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$
answered Nov 23 '18 at 7:22
KM101KM101
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
add a comment |
$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$
$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$
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 $1-cos^2theta = sin^2 theta$.
$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$
answered Nov 23 '18 at 7:22
KM101KM101
5,7411423
$$frac{sec^2 theta-cos^2 theta}{tan^2 theta}$$
$$=frac{frac{1}{cos^2 theta}-cos^2 theta}{tan^2 theta}$$
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 $1-cos^2theta = sin^2 theta$.
$$= frac{{(sin^2 theta)(1+cos^2 theta)}}{sin^2 theta} = 1+cos^2 theta$$
answered Nov 23 '18 at 7:22
KM101KM101
5,7411423
answered Nov 23 '18 at 7:22
KM101KM101
5,7411423
answered Nov 23 '18 at 7:22
KM101KM101
5,7411423
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
add a comment |
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
add a comment |
For $cos theta neq 0$ we have
$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$
answered Nov 23 '18 at 7:17
gimusigimusi
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
add a comment |
For $cos theta neq 0$ we have
$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$
answered Nov 23 '18 at 7:17
gimusigimusi
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
add a comment |
For $cos theta neq 0$ we have
$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$
answered Nov 23 '18 at 7:17
gimusigimusi
1
For $cos theta neq 0$ we have
$$frac {cos^2theta}{cos^2theta}cdot frac {frac {1}{cos^2theta}- cos^2theta} {tan^2theta}=frac{1-cos^4 theta}{sin^2theta}=frac{(1-cos^2 theta)(1+cos^2 theta)}{1-cos^2 theta}=1+cos^2 theta$$
answered Nov 23 '18 at 7:17
gimusigimusi
1
answered Nov 23 '18 at 7:17
gimusigimusi
1
answered Nov 23 '18 at 7:17
gimusigimusi
1
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
add a comment |
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
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@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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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