1. ## Trig identity proof

prove that $\displaystyle tan50^{o} -tan40^{o} = 2tan10^{o}$

2. Use these formulas:

$\displaystyle \tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}$

and $\displaystyle \cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]$

3. Originally Posted by red_dog
Use these formulas:

$\displaystyle \tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}$

and $\displaystyle \cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]$
thanks,

but I have never seen these formula's or been taught them. where did you get them from?

surely If I needed to use these formula's It would be in the book, that the question is from?

4. Originally Posted by Tweety
thanks,

but I have never seen these formula's or been taught them. where did you get them from?

surely If I needed to use these formula's It would be in the book, that the question is from?

Prove that $\displaystyle \tan 50 - \tan 40 = 2\tan 10$
Hi Tweety,

I don't know what formulas you have been taught, but here's another proof using the Tan sum formula:

$\displaystyle \tan(u+v)=\frac{\tan u+\tan v}{1-\tan u \tan v}$.

$\displaystyle \tan (50) = \tan (40+10)= \frac{\tan 40 + \tan 10}{1- \tan 40 \tan 10}$

$\displaystyle \tan 50 (1- \tan 40 \tan 10) = \tan 40 + \tan 10$

Now, distribute:

$\displaystyle \tan 50 - {\color{red}\tan 50 \tan 40} \tan 10 = \tan 40 + \tan 10$

You know that $\displaystyle {\color{red}\tan 50 \tan 40 = 1}$

$\displaystyle \tan 50 - \tan 10= \tan 40 + \tan 10$

$\displaystyle \tan 50 = \tan 40 + 2 \tan 10$

$\displaystyle \tan 50 - \tan 40 = 2 \tan 10$