Question

Question #d64b4

Answer 1

Please refer to the Explanation for the Proof.

Explanation:

Let `arc tan(1/2)=alpha, arc tan(1/5)=beta, and, arc tan(1/8)=gamma`.

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

Recall that,

`arc tan theta =x, x in RR iff tan theta=x, theta in (-pi/2,pi/2).`

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

Now, `tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)`

`=(1/2+1/5)/(1-1/10)=(7/10)/(9/10)=7/9, and, is +ve, so, alpha+beta in (0,pi/2)`.

Call `alpha+beta=delta," so, "tandelta=7/9, delta in (0,pi/2).`

Finally, `tan(gamma+delta)=(tangamma+tandelta)/(1-tangammatandelta)`

`=(1/8+7/9)/(1-7/72)=(65/72)/(65/72)=1`

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

`:. gamma+delta=pi/4`

`rArr alpha+beta+gamma=pi/4, or,`

`arc tan(1/2)+arc tan(1/5)+arc tan(1/8)=pi/4`.

Answer 2

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)`

`arc tan(frac(1/2+1/5) (1-1/10))+arc tan(1/8)...[because,(1/2)(1/5)=1/10lt 1]`

`=arc tan(7/9)+arc tan(1/8)`

`=arc tan (frac (7/9+1/8) (1-7/72))...[because,(7/9)(1/8)=7/72lt1]`

`=arc tan(frac(65/72)(65/72))`

`=arc tan1`

`=pi/4`

Hence the Proof.