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Find the minimum value of $a tan^2 x + b cot^2 x,$ where $text a is greater than b, b is greater than 0.$

more than one minimum value?Find the minimum value of $|sin x+cos x+tan x+cot x+sec x+csc x|$ for real numbers $x$Find the value of $tan^2alpha+cot^2beta$Find the minimum value of $sin^2 theta +cos^2 theta+sec^2 theta+csc^2 theta+tan^2 theta+cot^2 theta$The angle giving minimum valueIf $x + y + z + w=5$ then the minimum value of $x^2 cot (9°) + y^2 cot (27°) + z^2 cot (63°) + w^2 cot (81°)$ is?Find the value of $cos tan^-1 sin cot^-1 (x)$ .Finding the minimum value of $cot^2A + cot^2B+ cot^2C$ where $A$, $B$ and $C$ are angles of a triangle.Find the value of $cot(16)cot(44)+cot(44)cot(76)-cot(76)cot(16)$Find the value of $cot^-121+cot^-113+cot^-1(-8)$

0

$begingroup$

Find the minimum value of $a tan^2 x + b cot^2 x.text a is greater than b, b is greater than 0$.
Closest thing to solution I can come up with is this
$(sqrta tan x + sqrtbcot x )^2 -2 sqrtasqrtb $

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $tan(x)=t$, you can find the minimum value of $at^2+b/t^2$ where $t in BbbR$.
  $endgroup$
  – Xin Fu
  Mar 30 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @XinFu hmm then what ?
  $endgroup$
  – swarnim
  Mar 30 at 4:49
  

  
  
  
  

add a comment | 

0

$begingroup$

Find the minimum value of $a tan^2 x + b cot^2 x.text a is greater than b, b is greater than 0$.
Closest thing to solution I can come up with is this
$(sqrta tan x + sqrtbcot x )^2 -2 sqrtasqrtb $

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $tan(x)=t$, you can find the minimum value of $at^2+b/t^2$ where $t in BbbR$.
  $endgroup$
  – Xin Fu
  Mar 30 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @XinFu hmm then what ?
  $endgroup$
  – swarnim
  Mar 30 at 4:49
  

  
  
  
  

add a comment | 

$begingroup$

Find the minimum value of $a tan^2 x + b cot^2 x.text a is greater than b, b is greater than 0$.
Closest thing to solution I can come up with is this
$(sqrta tan x + sqrtbcot x )^2 -2 sqrtasqrtb $

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

$endgroup$

share|cite|improve this question

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

share|cite|improve this question

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

edited Mar 30 at 4:51

Robert Lewis

48.9k23168

asked Mar 30 at 4:38

swarnimswarnim

938

asked Mar 30 at 4:38

swarnimswarnim

938

3 Answers
3

active

oldest

votes

1

$begingroup$

Let $t=(tan , x)^2$. Note that $at+frac b t =(sqrt at -sqrt frac b t)^2+2sqrt ab geq 2sqrt ab$. Also equality holds when $t=sqrt b/a$. Hence the minimum value is $2sqrt ab$ which is attained when $tan , x=(b/a)^1/4$.

share|cite|improve this answer

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

add a comment | 

1

$begingroup$

The AM-GM inequality says: $atan^2x + bcot^2x ge 2sqrtatan^2xcdot bcot^2x= 2sqrtab$. What does this mean...?

share|cite|improve this answer

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

0

$begingroup$

You are almost there:

$(√a|tan x| -√b|cot x|)^2 +$

$2√a√b|tan x||cot x| ge$

$ 2√a√b|tan x cot x| =2√a√b.$

Equality:

$√a|tan x| -√b|cot x| =0.$

$tan ^2 x =√b/√a.$

share|cite|improve this answer

edited Mar 30 at 12:14

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

$endgroup$

add a comment | 

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1

$begingroup$

Let $t=(tan , x)^2$. Note that $at+frac b t =(sqrt at -sqrt frac b t)^2+2sqrt ab geq 2sqrt ab$. Also equality holds when $t=sqrt b/a$. Hence the minimum value is $2sqrt ab$ which is attained when $tan , x=(b/a)^1/4$.

share|cite|improve this answer

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

add a comment | 

1

$begingroup$

Let $t=(tan , x)^2$. Note that $at+frac b t =(sqrt at -sqrt frac b t)^2+2sqrt ab geq 2sqrt ab$. Also equality holds when $t=sqrt b/a$. Hence the minimum value is $2sqrt ab$ which is attained when $tan , x=(b/a)^1/4$.

share|cite|improve this answer

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

add a comment | 

$begingroup$

Let $t=(tan , x)^2$. Note that $at+frac b t =(sqrt at -sqrt frac b t)^2+2sqrt ab geq 2sqrt ab$. Also equality holds when $t=sqrt b/a$. Hence the minimum value is $2sqrt ab$ which is attained when $tan , x=(b/a)^1/4$.

share|cite|improve this answer

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

share|cite|improve this answer

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

edited Mar 30 at 5:37

trancelocation

13.7k1829

edited Mar 30 at 5:37

trancelocation

13.7k1829

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

answered Mar 30 at 4:51

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

1

$begingroup$

The AM-GM inequality says: $atan^2x + bcot^2x ge 2sqrtatan^2xcdot bcot^2x= 2sqrtab$. What does this mean...?

share|cite|improve this answer

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

1

$begingroup$

The AM-GM inequality says: $atan^2x + bcot^2x ge 2sqrtatan^2xcdot bcot^2x= 2sqrtab$. What does this mean...?

share|cite|improve this answer

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

$begingroup$

The AM-GM inequality says: $atan^2x + bcot^2x ge 2sqrtatan^2xcdot bcot^2x= 2sqrtab$. What does this mean...?

share|cite|improve this answer

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

$endgroup$

share|cite|improve this answer

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

answered Mar 30 at 4:51

DeepSeaDeepSea

71.4k54488

0

$begingroup$

You are almost there:

$(√a|tan x| -√b|cot x|)^2 +$

$2√a√b|tan x||cot x| ge$

$ 2√a√b|tan x cot x| =2√a√b.$

Equality:

$√a|tan x| -√b|cot x| =0.$

$tan ^2 x =√b/√a.$

share|cite|improve this answer

edited Mar 30 at 12:14

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

$endgroup$

add a comment | 

0

$begingroup$

You are almost there:

$(√a|tan x| -√b|cot x|)^2 +$

$2√a√b|tan x||cot x| ge$

$ 2√a√b|tan x cot x| =2√a√b.$

Equality:

$√a|tan x| -√b|cot x| =0.$

$tan ^2 x =√b/√a.$

share|cite|improve this answer

edited Mar 30 at 12:14

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

$endgroup$

add a comment | 

$begingroup$

You are almost there:

$(√a|tan x| -√b|cot x|)^2 +$

$2√a√b|tan x||cot x| ge$

$ 2√a√b|tan x cot x| =2√a√b.$

Equality:

$√a|tan x| -√b|cot x| =0.$

$tan ^2 x =√b/√a.$

share|cite|improve this answer

edited Mar 30 at 12:14

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

$endgroup$

share|cite|improve this answer

edited Mar 30 at 12:14

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822

answered Mar 30 at 5:17

Peter SzilasPeter Szilas

11.9k2822