$={\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+{\sin }^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\dots \dots +{\sin }^{-1}\left(\frac{\sqrt{\mathrm{n}}-\sqrt{\mathrm{n}-1}}{\sqrt{\mathrm{n}(\mathrm{n}+1)}}\right)$

Let $\mathrm{S}={\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+{\sin }^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\dots \dots +{\sin }^{-1}\left(\frac{\sqrt{\mathrm{n}}-\sqrt{\mathrm{n}-1}}{\sqrt{\mathrm{n}(\mathrm{n}+1)}}\right)$

Now, ${\mathrm{T}}_{\mathrm{n}}={\sin }^{-1}\left(\frac{\sqrt{\mathrm{n}}-\sqrt{\mathrm{n}-1}}{\sqrt{\mathrm{n}(\mathrm{n}+1)}}\right)$

$\begin{array}{l}={\sin }^{-1}\left[\frac{1}{\sqrt{\mathrm{n}}}\sqrt{1-{\left(\frac{1}{\sqrt{\mathrm{n}+1}}\right)}^2}-\frac{1}{\sqrt{\mathrm{n}+1}}\sqrt{1-{\left(\frac{1}{\sqrt{\mathrm{n}}}\right)}^2}\right]\\ ={\sin }^{-1}\frac{1}{\sqrt{\mathrm{n}}}-{\sin }^{-1}\frac{1}{\sqrt{\mathrm{n}+1}}\\ \therefore \mathrm{S}={\sin }^{-1}\frac{1}{\sqrt{2}}+\left({\sin }^{-1}\frac{1}{\sqrt{2}}-\sin \frac{1}{\sqrt{3}}\right)+\left({\sin }^{-1}\frac{1}{\sqrt{3}}-{\sin }^{-1}\frac{1}{\sqrt{4}}\right)+\dots ..+\mathrm{\infty }\\ =2{\sin }^{-1}\frac{1}{\sqrt{2}}=2\left(\frac{\mathrm{\pi }}{4}\right)=\frac{\mathrm{\pi }}{2}\end{array}$