Mryntu

Placing subfig vertically

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If $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$ [on hold]

help me to show that $fractan(alpha-beta)+tanbeta1-tan(alpha-beta)tanbeta=fracm^2-n^22mn$If $alpha = frac2pi7$ then the find the value of $tanalpha .tan2alpha +tan2alpha tan4alpha +tan4alpha tanalpha.$Find $tan(alpha+beta)$ when $tan beta = fracnsinalphacosalpha1-ncos^2alpha$.If $fracm+1m-1=fraccos(alpha-beta)sin(alpha+beta)$„ thenShow that $tan alpha tan beta + tan betatan gamma +tan gammatan alpha =1$If $alpha+beta+gamma=fracpi2,$ prove a expression equal to $fracsinalpha+sinbeta+singamma-1cosalpha+cosbeta+cosgamma$If $cosalpha = frac2cosbeta - 12-cosbeta$ , $(0<alpha , beta< pi)$, then $tanfracalpha2cotfracbeta2$ is equal to?If $theta=alpha+beta$ such that $fractanalphatanbeta=fracxy$, then $sin(alpha-beta)=fracx-yx+ysintheta$Given $tanbeta=fracnsinalphacosalpha1-nsin^2alpha$, show that $tan(alpha-beta)=(1-n)tanalpha$Solving for $beta$ in $frac mtan(alpha-beta) cos^2beta = frac ntanbetacos^2(alpha-beta)$

1

1

$begingroup$

Given that $$tan(A+B)= fractan A + tan B1-tan A tan B$$

Show that if $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$.

trigonometry

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edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

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New contributor

Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota 2 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
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  Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
  $endgroup$
  – Martin Hansen
  13 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
  $endgroup$
  – Barry Cipra
  13 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does your question insist on the identity being used ?
  $endgroup$
  – Martin Hansen
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
  $endgroup$
  – Nour G
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

1

1

$begingroup$

Given that $$tan(A+B)= fractan A + tan B1-tan A tan B$$

Show that if $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$.

trigonometry

share|cite|improve this question

edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

143

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Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota 2 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
  $endgroup$
  – Martin Hansen
  13 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
  $endgroup$
  – Barry Cipra
  13 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does your question insist on the identity being used ?
  $endgroup$
  – Martin Hansen
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
  $endgroup$
  – Nour G
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Given that $$tan(A+B)= fractan A + tan B1-tan A tan B$$

Show that if $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$.

trigonometry

share|cite|improve this question

edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

143

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Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this question

edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

143

New contributor

Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

143

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Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 7 hours ago

Blue

49k870156

edited 7 hours ago

Blue

49k870156

asked 13 hours ago

Nour GNour G

143

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asked 13 hours ago

Nour GNour G

143

5 Answers
5

active

oldest

votes

1

$begingroup$

Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac2tan alpha1-tan^2 alpha$$
and
$$tan(2 beta)=frac2tan beta1-tan^2 beta$$
Thus
$$fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$

Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$

Observe that
$$tan alpha=fracxy$$
and
$$tan beta=fracyx$$
Thus
$$LHS=fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$

share|cite|improve this answer

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm Thanks for the suggested edit
  $endgroup$
  – Martin Hansen
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

If

$$2alpha+2beta=180°$$

then

$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.

share|cite|improve this answer

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
  $endgroup$
  – Martin Hansen
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Makes so much sense now.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.

Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$

Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

VasyaVasya

4,0881618

$endgroup$

add a comment | 

0

$begingroup$

$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$

share|cite|improve this answer

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

$endgroup$

add a comment | 

0

$begingroup$

$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$

Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

saket kumarsaket kumar

42111

$endgroup$

add a comment | 

1

$begingroup$

Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac2tan alpha1-tan^2 alpha$$
and
$$tan(2 beta)=frac2tan beta1-tan^2 beta$$
Thus
$$fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$

Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$

Observe that
$$tan alpha=fracxy$$
and
$$tan beta=fracyx$$
Thus
$$LHS=fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$

share|cite|improve this answer

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm Thanks for the suggested edit
  $endgroup$
  – Martin Hansen
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac2tan alpha1-tan^2 alpha$$
and
$$tan(2 beta)=frac2tan beta1-tan^2 beta$$
Thus
$$fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$

Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$

Observe that
$$tan alpha=fracxy$$
and
$$tan beta=fracyx$$
Thus
$$LHS=fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$

share|cite|improve this answer

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm Thanks for the suggested edit
  $endgroup$
  – Martin Hansen
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac2tan alpha1-tan^2 alpha$$
and
$$tan(2 beta)=frac2tan beta1-tan^2 beta$$
Thus
$$fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$

Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$

Observe that
$$tan alpha=fracxy$$
and
$$tan beta=fracyx$$
Thus
$$LHS=fractan(2 alpha)tan(2 beta)=frac2tan alpha(1-tan^2 beta)(1-tan^2 alpha) 2tan beta$$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$

share|cite|improve this answer

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

$endgroup$

share|cite|improve this answer

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

edited 8 hours ago

Teepeemm

69459

edited 8 hours ago

Teepeemm

69459

answered 13 hours ago

Martin HansenMartin Hansen

27713

answered 13 hours ago

Martin HansenMartin Hansen

27713

4

$begingroup$

If

$$2alpha+2beta=180°$$

then

$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.

share|cite|improve this answer

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
  $endgroup$
  – Martin Hansen
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Makes so much sense now.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

If

$$2alpha+2beta=180°$$

then

$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.

share|cite|improve this answer

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
  $endgroup$
  – Martin Hansen
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Makes so much sense now.
  $endgroup$
  – Nour G
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If

$$2alpha+2beta=180°$$

then

$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.

share|cite|improve this answer

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

$endgroup$

share|cite|improve this answer

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

answered 13 hours ago

Yves DaoustYves Daoust

130k676227

2

$begingroup$

Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.

Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$

Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

VasyaVasya

4,0881618

$endgroup$

add a comment | 

2

$begingroup$

Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.

Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$

Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

VasyaVasya

4,0881618

$endgroup$

add a comment | 

$begingroup$

Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.

Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$

Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

VasyaVasya

4,0881618

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

VasyaVasya

4,0881618

answered 13 hours ago

VasyaVasya

4,0881618

answered 13 hours ago

VasyaVasya

4,0881618

0

$begingroup$

$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$

share|cite|improve this answer

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

$endgroup$

add a comment | 

0

$begingroup$

$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$

share|cite|improve this answer

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

$endgroup$

add a comment | 

$begingroup$

$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$

share|cite|improve this answer

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

$endgroup$

share|cite|improve this answer

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

answered 13 hours ago

Maria MazurMaria Mazur

46.6k1260119

0

$begingroup$

$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$

Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

saket kumarsaket kumar

42111

$endgroup$

add a comment | 

0

$begingroup$

$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$

Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

saket kumarsaket kumar

42111

$endgroup$

add a comment | 

$begingroup$

$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$

Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

saket kumarsaket kumar

42111

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

saket kumarsaket kumar

42111

answered 2 hours ago

saket kumarsaket kumar

42111

answered 2 hours ago

saket kumarsaket kumar

42111