Proving tangent

[Diego14]

show the sequence of steps to prove the addition formula for tangent
tan(s+t)=tan s + tan t
1-tan^2 s tan^2 t

first i LHS=tan s + tan t
1-tan^2 s tan^2 t
then i multiplied this with the Subtraction formula for tangent
tan s + tan t (tan s-tan t)
1-tan^2 s tan^2 t (1+tan^2 s tan^2 t)
that got me
tan^2 s- tan^2t
???? i didnt know how to multiply the two--am i in the right direction or did i miss it :?: :?:

 

[Diego14]

anyone? Please?

 

[Mrspi]

show the sequence of steps to prove the addition formula for tangent
tan(s+t)=tan s + tan t
1-tan^2 s tan^2 t

first i LHS=tan s + tan t
1-tan^2 s tan^2 t
then i multiplied this with the Subtraction formula for tangent
tan s + tan t (tan s-tan t)
1-tan^2 s tan^2 t (1+tan^2 s tan^2 t)
that got me
tan^2 s- tan^2t
???? i didnt know how to multiply the two--am i in the right direction or did i miss it :?: :?:

Look here: http://www.intmath.com/Analytic-trigono ... angles.php

for a nice proof!

 

[chrisr]

Hi Diego

tanA = SinA/{CosA}

tan{A+B} = Sin{A+B}/Cos{A+B}

Now use the identities for these

{SinACosB + CosASinB}/{CosACosB - SinASinB}

Now multiply by 1, in the form of {1/CosB}/{1/CosB},
in other words, divide numerator and denominator by CosB, to get

{(SinACosB/CosB) + (CosASinB/CosB)} divided by {(CosACosB/CosB) - (SinASinB/CosB)}

which is {SinA + CosAtanB}/{CosA-SinAtanB}

Finally, multiply by 1 again, this time in the form of (1/CosA)/(1/CosA)
in other words divide numerator and denominator by CosA

Then you have the answer.
That's the way I would solve problems like that.

Check the link given by Mrspi in case it's simpler