Thread: one more proof

in this problem i have to prove

the inverse of cot (2) + inverse of cot (3) = pi/4

so i rewrote the problem in terms of inverse tan and got

inverse of tan (1/2) + inverse of tan (1/3)

after i get to this step im not sure where to go...i used triangles trying to figure out the angle measures and came up with some weird numbers...one triangle showed sin = 1/sqrt of 5 and cos = 2/sqrt 5. the other triangle shows sin x = 1/sqrt of 10 and cos x = 3/sqrt of 10? any help would be appreciated. thanks in advance

Hi

$\displaystyle \tan \left(\tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right)\right) = \frac{\tan \left(\tan^{-1}\left(\frac12\right)\right) + \tan \left(\tan^{-1}\left(\frac13\right)\right)}{1 - \tan \left(\tan^{-1}\left(\frac12\right)\right) \times \tan \left(\tan^{-1}\left(\frac13\right)\right)}$

$\displaystyle \tan \left(\tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right)\right) = \frac{\frac12 + \frac13}{1 - \frac12 \times \frac13}$

$\displaystyle \tan \left(\tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right)\right) = 1 = \tan\left(\frac{\pi}{4}\right)$

$\displaystyle 0 < \tan^{-1}\left(\frac12\right) < \frac{\pi}{2}$

$\displaystyle 0 < \tan^{-1}\left(\frac13\right) < \frac{\pi}{2}$

$\displaystyle 0 < \tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right) < \pi$

Therefore $\displaystyle \tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right) = \frac{\pi}{4}$