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 $frac mtan(alpha-beta) cos^2beta = frac ntanbetacos^2(alpha-beta)$
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 $frac mtan(alpha-beta) cos^2beta = frac ntanbetacos^2(alpha-beta)$
$begingroup$
Given that $$tan(A+B)= fractan A + tan B1-tan A tan B$$
Show that if $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$.
trigonometry
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
|
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
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
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
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given that $$tan(A+B)= fractan A + tan B1-tan A tan B$$
Show that if $alpha+beta=90^circ$, then $fractan2alphatan2beta =-1$.
trigonometry
trigonometry
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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
![](https://i.stack.imgur.com/zvoFI.png?s=32&g=1)
![](https://i.stack.imgur.com/zvoFI.png?s=32&g=1)
Blue
49k870156
49k870156
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
![](https://lh6.googleusercontent.com/-DeWHNsA7Ee4/AAAAAAAAAAI/AAAAAAAAHEw/dRDDOLFMTZQ/photo.jpg?sz=32)
![](https://lh6.googleusercontent.com/-DeWHNsA7Ee4/AAAAAAAAAAI/AAAAAAAAHEw/dRDDOLFMTZQ/photo.jpg?sz=32)
Nour GNour G
143
143
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
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
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
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
3
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
$begingroup$
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
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 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
$begingroup$
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
$begingroup$
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
$begingroup$
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
5 Answers
5
active
oldest
votes
$begingroup$
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 $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$
$endgroup$
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 |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.
$endgroup$
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 |
$begingroup$
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)
$endgroup$
add a comment |
$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
$endgroup$
add a comment |
$begingroup$
$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$
$endgroup$
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 $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$
$endgroup$
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 $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$
$endgroup$
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 $x^2y^2$
$$=fracfrac2xybig(1-fracy^2x^2big)big(1-fracx^2y^2big) frac2yx times fracx^2y^2x^2y^2$$
$$=frac2xy(x^2-y^2)(y^2-x^2)2xy$$
$$=-1$$
$$=RHS$$
edited 8 hours ago
Teepeemm
69459
69459
answered 13 hours ago
Martin HansenMartin Hansen
27713
27713
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
$begingroup$
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
1
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
$begingroup$
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
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
8 hours ago
add a comment |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.
$endgroup$
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.
$endgroup$
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.
$endgroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=fractan2alpha+tan2beta1-tan2alphatan2beta=0$$ and the claim easily follows.
answered 13 hours ago
Yves DaoustYves Daoust
130k676227
130k676227
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 |
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
1
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
$begingroup$
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
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Thank you so much! Makes so much sense now.
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– Nour G
12 hours ago
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Thank you so much! Makes so much sense now.
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– Nour G
12 hours ago
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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)
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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)
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add a comment |
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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)
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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)
edited 13 hours ago
answered 13 hours ago
VasyaVasya
4,0881618
4,0881618
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$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
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$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
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add a comment |
$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
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$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered 13 hours ago
![](https://i.stack.imgur.com/StKep.png?s=32&g=1)
![](https://i.stack.imgur.com/StKep.png?s=32&g=1)
Maria MazurMaria Mazur
46.6k1260119
46.6k1260119
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$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$
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$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$
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$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$
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$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$fractan2alphatan2beta =fractan2alphatan(180 ^circ-2alpha) =fractan2alpha-tan2alpha = -1= RHS $$
edited 2 hours ago
answered 2 hours ago
![](https://lh5.googleusercontent.com/-z-NV9eL8IyE/AAAAAAAAAAI/AAAAAAAAAGY/0CEHr5YAqLE/photo.jpg?sz=32)
![](https://lh5.googleusercontent.com/-z-NV9eL8IyE/AAAAAAAAAAI/AAAAAAAAAGY/0CEHr5YAqLE/photo.jpg?sz=32)
saket kumarsaket kumar
42111
42111
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3
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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.
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– saulspatz
13 hours ago
1
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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
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– Martin Hansen
13 hours ago
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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 $180^circ$.
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– Barry Cipra
13 hours ago
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Does your question insist on the identity being used ?
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– Martin Hansen
13 hours ago
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Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
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– Nour G
13 hours ago