Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Set $\theta =(1/2){\cos }^{-1}(x^2)$, with 0≤$\theta <\pi /4$. Then $\sqrt{1+x^2}=\sqrt{2}cos$ θ and $\sqrt{1-x^2}=\sqrt{2}sin$ θ. The ratio is $\tan (\pi /4+\theta )$, whose angle lies in [$\pi /4,\pi /2$), inside the principal tan⁻¹ range. Hence the answer is $\pi /4+(1/2){\cos }^{-1}(x^2)$.