# Question d64b4

Jan 26, 2017

Please refer to the Explanation for the Proof.

#### Explanation:

Let $a r c \tan \left(\frac{1}{2}\right) = \alpha , a r c \tan \left(\frac{1}{5}\right) = \beta , \mathmr{and} , a r c \tan \left(\frac{1}{8}\right) = \gamma$.

:. tanalpha=1/2, tanbeta=1/5, &, tangamma=1/8.

Recall that,

$a r c \tan \theta = x , x \in \mathbb{R} \iff \tan \theta = x , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

Since, all tanalpha,tanbeta" &, "tangamma gt 0; alpha,beta,gamma in (0,pi/2).

Now, $\tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

$= \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{10}} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{9} , \mathmr{and} , i s + v e , s o , \alpha + \beta \in \left(0 , \frac{\pi}{2}\right)$.

Call $\alpha + \beta = \delta , \text{ so, } \tan \delta = \frac{7}{9} , \delta \in \left(0 , \frac{\pi}{2}\right) .$

Finally, $\tan \left(\gamma + \delta\right) = \frac{\tan \gamma + \tan \delta}{1 - \tan \gamma \tan \delta}$

$= \frac{\frac{1}{8} + \frac{7}{9}}{1 - \frac{7}{72}} = \frac{\frac{65}{72}}{\frac{65}{72}} = 1$

Here, delta, gamma in (0,pi/2) rArr gamma+delta in (0,pi); & because, tan(gamma+delta)=1,

$\therefore \gamma + \delta = \frac{\pi}{4}$

$\Rightarrow \alpha + \beta + \gamma = \frac{\pi}{4} , \mathmr{and} ,$

$a r c \tan \left(\frac{1}{2}\right) + a r c \tan \left(\frac{1}{5}\right) + a r c \tan \left(\frac{1}{8}\right) = \frac{\pi}{4}$.

Jan 26, 2017

See the Proof in Explanation.

#### Explanation:

As a Second Method, we can solve the Problem by using the following Useful Result :

arc tanx+arc tany=arc tan(frac(x+y)(1-xy)); x,y in RR^+ xy lt 1.

"The L.H.S.={arc tan(1/2)+arc tan(1/5)}+arctan(1/8)#

$a r c \tan \left(\frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{10}}\right) + a r c \tan \left(\frac{1}{8}\right) \ldots \left[\because , \left(\frac{1}{2}\right) \left(\frac{1}{5}\right) = \frac{1}{10} < 1\right]$

$= a r c \tan \left(\frac{7}{9}\right) + a r c \tan \left(\frac{1}{8}\right)$

$= a r c \tan \left(\frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{72}}\right) \ldots \left[\because , \left(\frac{7}{9}\right) \left(\frac{1}{8}\right) = \frac{7}{72} < 1\right]$

$= a r c \tan \left(\frac{\frac{65}{72}}{\frac{65}{72}}\right)$

$= a r c \tan 1$

$= \frac{\pi}{4}$

Hence the Proof.