Question

Solve for $x$:${\sin }^{-1}\frac{x}{\sqrt{1+x^2}}+{\sin }^{-1}\frac{1}{\sqrt{1+x^2}}={\sin }^{-1}\left(\frac{1+x}{\sqrt{1+x^2}}\right)$

Answer 1

${\sin }^{-1}\frac{x}{\sqrt{1+x^2}}+{\sin }^{-1}\frac{1}{\sqrt{1+x^2}}={\sin }^{-1}\left(\frac{1+x}{\sqrt{1+x^2}}\right)\Rightarrow {\sin }^{-1}(\frac{x}{\sqrt{1+x^2}}\sqrt{1-{\left(\frac{1}{\sqrt{1+x^2}}\right)}^2}+\frac{1}{\sqrt{1+x^2}}\sqrt{1-{\left(\frac{x}{\sqrt{1+x^2}}\right)}^2})\Rightarrow {\sin }^{-1}(\frac{x}{\sqrt{1+x^2}}\sqrt{\frac{1+x^2-1}{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\sqrt{\frac{1+x^2-x^2}{1+x^2}})\Rightarrow {\sin }^{-1}(\frac{x}{\sqrt{1+x^2}}\frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+x^2}})\Rightarrow {\sin }^{-1}\left(\frac{1+x^2}{1+x^2}\right)\Rightarrow {\sin }^{-1}\left(1\right)\Rightarrow \frac{\pi }{2}$