Question

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

Answer 1

`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`