Prove:-

arctan(1/4)+ arctan(1/9)= (1/2)arccos(3/5)

2 answers

Using the identity arctanA + arctanB = arctan((A+B)/(1-AB)),

arctan(1/4)+ arctan(1/9)
= arctan((1/4 + 1/9)/(1 - 1/36))
= arctan(13/35)

Now,

Using the identity
arctanx = (1/2)arccos((1-x^2)/(1+x^2))
=> arctan(13/35) = (1/2)arccos(1-(13/35)^2/1+(13/35)^2)
= (1/2)arccos(35^2-13^2/35^2+13^2)
= (1/2)arccos(1225-169/1225+169)
= (1/2)arccos(1056/1394)

I'm getting a slightly different answer. Perhaps one of the tutors could proofread my result
Maybe it's because the equation is false.

arctan(1/4)+arctan(1/9) = 0.3556
(1/2)arccos(3/5) = 0.4636
Similar Questions
  1. Now we prove Machin's formula using the tangent addition formula:tan(A+B)= tanA+tanB/1-tanAtanB. If A= arctan(120/119) and B=
    1. answers icon 2 answers
    1. answers icon 1 answer
  2. 4 arctan(1/3)+4 arctan(1/4)+4 arctan(2/9)=pihow do i prove it but without using this formula arctan(x)+arctan(y) = arctan(
    1. answers icon 1 answer
  3. arctan(tan(2pi/3)thanks. arctan(tan(2pi/3) = -pi/3 since arctan and tan are inverse operations, the solution would be 2pi/3 the
    1. answers icon 0 answers
more similar questions