How do you evaluate #arctan((arctan(9/7) - arctan(7/6)) / (arctan(5/3) - arctan(3/2)))#?

1 Answer
Apr 1, 2016

#pi/4+K*pi, K in NN#

Explanation:

Let's rewrite the expression as

#epsilon=arctan(theta/rho)#
with
#theta=arctan(9/7)-arctan(7/6)# => #tan theta=tan(arctan(9/7)-arctan(7/6))#
#rho=arctan(5/3)-arctan(3/2)# => #tan rho=tan (arctan(5/3)-arctan(3/2))#

Using the formula of the tangent of the difference between two angles
#tan(x-y)=(tanx-tany)/(1+tanx*tany)#
we get

#tan theta=(9/7-7/6)/(1+9/cancel(7)*cancel(7)/6)=((54-49)/42)/(1+3/2)=2/cancel5*cancel5/42=1/21#
#-> theta=arctan(1/21)#
and
#tan rho=(5/3-3/2)/(1+5/cancel3*cancel3/2)=((10-9)/6)/(1+5/2)=2/7*1/6=1/21#
#-> rho=arctan(1/21)#

So
#epsilon=arctan(theta/rho)=arctan(cancel(arctan(1/21))/cancel(arctan(1/21)))=arctan(1)#
=> #epsilon=pi/4+K*pi, K in NN#