Solve tan^2a=tanbtanc: Prove and Simplify

[shravan]

sin(a+b)/sin(a+c)=[ sin(2b)/sin(2c)]^(1/2)
then prove tan^2a=tanbtanc
I have reached till {tan(a)cos(b)+ sin(b)} * {sin(c)cos(c)}^(1/2)=
{tan(a)cos(c)+sin(c)}* { sin(b) cos(b)}^(1/2)

 

[marlon]

sin(a+b)/sin(a+c)=[ sin(2b)/sin(2c)]^(1/2)
then prove tan^2a=tanbtanc
I have reached till {tan(a)cos(b)+ sin(b)} * {sin(c)cos(c)}^(1/2)=
{tan(a)cos(c)+sin(c)}* { sin(b) cos(b)}^(1/2)

Can you assume that a+b+c = 180 degrees ?

I think there is something missing in your question. How elese can you prove the first equality ?

Or is it given ? In that case, you can find a connection between the angles by manipulating this first equality.

marlon

 

[Gale]

you have all the info you need. i squared everything, and then solved for tan^2a. its messy, but it does work out. when you square it all, the tana terms cancel. and then you just have to factor and divide ect.

 

[shravan]

no there is no relation between the angles.the relation given is not going to help much. I think I will be getting a method for that sum.however thank u

 

[Gale]

it works out. i said that already. just keep trucking. like i said, square everything and start gathering terms. its just a lot of math, but it does work.

 

[NogardMX]

try using some of these identities

Tan=cos/sin

Sin(a+b)=
SinaCosb+CosaSinb

Cos(a+b)
CosaCosb-SinaSinb

ya that's my knowledge of trig

 

[Gale]

the OP already used those identities. the only other identy needed is knowing tanx= sinx/cosx. otherwise its all just rearranging the equation.