$\large \arctan (x)=2\arctan \left ( \frac{x}{1+\sqrt{1+x^{2}}} \right )$
$\large \arctan (x)=\int_{0}^{x}\frac{1}{z^{2}+1}dz\;;\;\left | x \right |\leq 1$
$\large \int \arctan (z)dz=z\arctan (z)-\frac{1}{2}\ln (1+z^{2})+C$
Arctangent formulas for $\pi$
$\large \frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$
$\large \frac{\pi }{4}=\arctan\frac{1}{2}+\arctan \frac{1}{3}$
$\large \frac{\pi }{4}=2\arctan\frac{1}{2}-\arctan \frac{1}{7}$
$\large \frac{\pi }{4}=2\arctan\frac{1}{3}+\arctan \frac{1}{7}$
$\large \frac{\pi }{4}=8\arctan\frac{1}{10}-4\arctan \frac{1}{515}-\arctan \frac{1}{239}$
$\large \frac{\pi }{4}=3\arctan\frac{1}{4}+\arctan \frac{1}{20}+\arctan \frac{1}{1985}$
$\large \frac{\pi }{4}=24\arctan\frac{1}{8}+8\arctan \frac{1}{57}+4\arctan \frac{1}{239}$
Question 1: Evaluate tan-1(1.732)?
Solution:
The given value is, tan-1(1.732)
From this given quantity, 1.732 can be written as a function of tan.
So, 1.732 = tan(60)
Therefore, tan-1(1.732) = tan-1 (tan(60)) = 60°
60° = 60 $\times$ $\frac{\pi}{180}$ = 1.047 radians.