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Answer 1 · 2017-01-26T14:18:01

Please refer to the Explanation for the Proof.

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 · 2017-01-26T14:40:29

See the Proof in 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.