Prove trig equation

[courteous]

Prove using addition theorem: \(\displaystyle \arctan({5})+\arctan({-\frac{2}{3})=\frac{\pi}{4}\)

My misthought $\displaystyle \rightarrow$ If you $\displaystyle \tan$ the whole equation, don't you get $\displaystyle 5-\frac{2}{3}=\tan{\frac{\pi}{4}}$ ? Guess not, as it disqualifies the equality, but what do you get :?:

 

[Deleted member 4993]

Prove using addition theorem: \(\displaystyle \arctan({5})+\arctan({-\frac{2}{3})=\frac{\pi}{4}\)

My misthought $\displaystyle \rightarrow$ If you $\displaystyle \tan$ the whole equation, don't you get $\displaystyle 5-\frac{2}{3}=\tan{\frac{\pi}{4}}$ ? Guess not, as it disqualifies the equality, but what do you get :?:

I t was not a misthought - almost good-thought (but almost)...

assume

$\displaystyle \arctan({5}) \, = \, \theta$

then

$\displaystyle \tan(\theta) \, = \, 5..........................................................(1)$

and

$\displaystyle \arctan({-\frac{2}{3}}) \, = \, \phi$

then

$\displaystyle \tan(\phi) \, = \, -\frac{2}{3}..................................................(2)$

then

You need to prove:

\(\displaystyle tan[(\arctan({5})+\arctan({-\frac{2}{3})] \, = \, tan[\frac{\pi}{4}]\)

or

$\displaystyle tan(\theta \, + \, \phi) \, = \, 1.................................................(3)$

Now continue....

 

[courteous]

... $\displaystyle \frac{5-\frac{2}{3}}{1+5*\frac{2}{3}}$