How do you evaluate #2((tan^-1(1/3))-tan^-1(-1/7)#?

1 Answer
Ratnaker Mehta
Jul 3, 2016

#pi/4#.

Explanation:

We have to use these Rules : #R(1) : tan^-1x+tan^-1y=tan^-1{(x+y)/(1-xy)}; x,y>0, xy<1#.

#R(2) : tan^-1(-x)=-(tan^-1x), x in RR#.

The Given Exp. #=2tan^-1(1/3)-tan^-1(-1/7)#
#=tan^-1(1/3)+tan^-1(1/3)+tan^-1(1/7).................[R(2)]#

#=tan^-1{(1/3+1/3)/(1-1/3*1/3)}+tan^-1(1/7)................[R(1)]#

#=tan^-1(2/3*9/8)+tan^-1(1/7)#
#tan^-1(3/4)+ tan^-1(1/7)#
#=tan^-1{(3/4+1/7)/(1-3/28)}................{R(1)]#
#=tan^-1{(25/28)/(25/28)}#
#=tan^-1(1)#
#=pi/4#.

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