How do you prove that #2\tan ^ { - 1} \frac { 1} { 5} + \tan ^ { - 1} \frac { 1} { 7} + 2\tan ^ { - 1} \frac { 1} { 8} = \frac { \pi } { 4}#?

2 Answers
Monzur R.
May 21, 2017

See full process below.

Explanation:

Let #2tan^-1(1/5)+tan^-1(1/7)+2tan^-1(1/8)=a#

Let #theta=tan^-1(1/5)# and #phi=tan^-1(1/7)# and #psi=tan^-1(1/8)#

Then #tantheta=1/5# and #tanphi=1/7# and #tanpsi=1/8#

#costheta=sqrt(1/(1+tan^2theta))=5/sqrt26#

#sintheta=sqrt(1-cos^2theta)=1/sqrt26#

#cosphi=sqrt(1/(1+tan^2phi))=7/sqrt50#

#sinphi=sqrt(1-cos^2phi)=1/sqrt50#

#cospsi=sqrt(1/(1+tan^2psi))=8/sqrt65#

#sinpsi=sqrt(1-cos^2psi)=1/sqrt65#

#2theta+phi+2psi=a#

Let #rho=2theta+2psi#

#phi+rho=a#

#sin(phi+rho)=sina#

#sin(phi+rho)#

#=sinphicosrho+cosphisinrho#

#=sinphi[cos(2theta+2psi)]+cosphi[sin(2theta+2psi)]#

#=sinphi(cos2thetacos2psi-sin2thetasin2psi)+cosphi(sin2thetacos2psi+cos2thetasin2psi)#

#=sinphi[(2cos^2theta-1)(2cos^2psi-1)-(2sinthetacostheta)(2sinpsicospsi)] + cosphi[(2sinthetacostheta)(2cos^2psi-1)+(2cos^2theta-1)(2sinpsicospsi)]#

#=1/sqrt50[(2(5/sqrt26)^2-1)(2(8/sqrt65)^2-1)-(2(1/sqrt26)(5/sqrt26))(2(1/sqrt65)(8/sqrt65))]#
#+7/sqrt50[(2(1/sqrt26)(5/sqrt26)(2(8/sqrt65)^2-1)+(2(5/sqrt26)^2-1)(2(1/sqrt65)(8/sqrt65))]=sqrt2/2#

#sin(phi+rho)=sina=sqrt2/2#

#phi+rho=a=pi/4+2kpi, k in ZZ#

#2theta+phi+2psi=a=pi/4+2kpi, k in ZZ#

# 0 < tantheta,tanphi,tanpsi < 1#

#therefore 0 < theta,phi,psi < pi/4#

#therefore 0 < 2theta+phi+2psi < (5pi)/4#

#therefore2theta+phi+2psi=pi/4#

#thereforea=pi/4#

#therefore2tan^-1(1/5)+tan^-1(1/7)+2tan^-1(1/8)=pi/4# #sf(QED)#

P dilip_k
May 22, 2017

#LHS=2tan^-1(1/5)+tan^-1(1/7)+2tan^-1(1/8)#

#=2(tan^-1(1/5)+tan^-1(1/8))+tan^-1 (1/7)#

#=2tan^-1((1/5+1/8)/(1-1/5xx1/8))+tan^-1(1/7)#

#=2tan^-1((13/40)/(39/40))+tan^-1(1/7)#

#=2tan^-1(1/3)+tan^-1(1/7)#

#=tan^-1((1/3+1/3)/(1-1/3xx1/3))+tan^-1(1/7)#

#=tan^-1((2/3)/(8/9))+tan^-1(1/7)#

#=tan^-1(3/4)+tan^-1(1/7)#

#=tan^-1((3/4+1/7)/(1-3/4xx1/7))#

#=tan^-1((25/28)/(25/28))#

#=tan^-1(1)=pi/4=RHS#

Related questions
Impact of this question
888 views around the world
You can reuse this answer
Creative Commons License