Let #varphi = arctan(3/4)#.

This means #tan varphi = 3/4#.

#tan varphi = 3/4 = sin varphi/cosvarphi = sin varphi/(+-sqrt(1-sin^2varphi)#

It is not specified which quadrant does the angle #varphi# is in, so we'll take the general case.

#sinvarphi/(+-sqrt(1-sin^2varphi)) = 3/4#

#=> sinvarphi = +-3/4 sqrt(1-sin^2varphi)#

Square both sides .

#sin^2 varphi = 9/16 (1-sin^2 varphi)#

#1*sin^2 varphi = 9/16 - 9/16 sin^2varphi#

#(1+9/16)sin^2varphi = 9/16#

#25/16 sin^2varphi = 9/16#

Now we can the square root of both :

#+-5/4 sin varphi = +-3/4 => sin varphi = +-(3/4 * 4/5)#

#sin varphi = +-3/5#.

Calculating #cos varphi#, you'll get

#cos varphi = +-4/5#

#color(red)("Important Note")#: #sin varphi# and #cos varphi# must have the same sign such that #tan varphi > 0#.

#=> varphi = arcsin(+-3/5) = arcsin (+-0.6)#

In order to simplify things, let's take only the positive value and assume #varphi in Q_I#.

#varphi^+ = arcsin(0.6)#

Notice how #0.6# is between #0.5# and #0.707 = sqrt2/2#.

We can deduce #varphi^+# to be in the interval #(pi/6,pi/4)#.

Considering how #sin varphi^+ ~~ (sin(pi/6) + sin(pi/4))/2#,

this implies that #varphi^+# is close to being in "between" #pi/6# and #pi/4#. In other words,

#varphi^+ ~~ (pi/6 +pi/4)/2#

#varphi^+ ~~ (5pi)/24#.

I know this answer has a lot of implied info, but it's as good as you can get without using a calculator or doing calculus.

The actual value of #sin((5pi)/24)# is ~#0.608#.

And finally, since the #tan# function has period #npi, n in ZZ#, this means that

#color(red)(arctan 0.75 in {(5pi)/24 + npi, n in ZZ}#.