How to solve this identity? Thank you!

Question

$\frac{tan (a + b) + tan (a - b)}{tan (a + b) - tan (a - b)} = \frac{sin (2 a)}{sin (2 b)}$ $(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)) = (sin(2a))/(sin(2b))$

1202 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-10T13:53:07

We know that, $tan a \pm tan b$ $tana+-tanb$

$= \frac{sin a}{cos a} \pm \frac{sin b}{cos b}$ $=sina/cosa+-sinb/cosb$

$= \frac{sin a cos b \pm cos a sin b}{cos a cos b}$ $=(sinacosb+-cosasinb)/(cosacosb$

$= \frac{sin (a \pm b)}{cos a \cdot cos b}$ $=sin(a+-b)/(cosa*cosb)$

So, $\frac{tan a + tan b}{tan a - tan b} = \frac{sin (a + b)}{sin (a - b)}$ $(tana+tanb)/(tana-tanb)=sin(a+b)/sin(a-b)$

$L H S = \frac{tan (a + b) + tan (a - b)}{tan (a + b) - tan (a - b)}$ $LHS=(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)$

$= \frac{sin (a + b + a - b)}{sin (a + b - (a - b))}$ $=sin(a+b+a-b)/(sin(a+b-(a-b))$

$= \frac{sin (2 a)}{sin (2 b)} = R H S$ $=sin(2a)/(sin(2b))=RHS$