Question

SOUTION Putting ( x=tan theta, ) we get
[
begin{aligned}
tan ^{-1}left(frac{sqrt{1+x^{2}}-1}{x}right) &=tan ^{-1}left(frac{sqrt{1+tan ^{2} theta}-1}{tan theta}right)
&=tan ^{-1}left(frac{sec theta-1}{tan theta}right)=tan ^{-1}left(frac{1-cos theta}{sin theta}right)
&=tan ^{-1}left{frac{2 sin ^{2}left(frac{theta}{2}right)}{2 sin left(frac{theta}{2}right) cos left(frac{theta}{2}right)}right}=tan ^{-1}left(tan frac{theta}{2}right)
therefore quad tan ^{-1}left(frac{sqrt{1+x^{2}}-1}{x}right) &=frac{theta}{2}=frac{1}{2} tan ^{-1} x
end{aligned}
]

# EXAMPLE 34 Prove that tan-1 V + *

Solution