Question

If ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\mathrm{\pi }$ then find the value of ${\left(\mathrm{x}+\mathrm{y}\right)}^2$.

Answer 1

Step 1: Compare the range of LHS and RHS

$$\begin{gathered} {\sin }^{-1}\left(x\right)\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\forall x\in \left[-1,1\right] \\ \Rightarrow -\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\left(x\right)\le \frac{\mathrm{\pi }}{2} \end{gathered}$$

Similarly,

$$\begin{gathered} {\sin }^{-1}\left(y\right)\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\forall y\in \left[-1,1\right] \\ \Rightarrow -\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\left(y\right)\le \frac{\mathrm{\pi }}{2} \end{gathered}$$.

Step 2: Find the value of ${\left(\mathrm{x}+\mathrm{y}\right)}^2$

Adding the above two inequations,

$-\mathrm{\pi }\le {\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)\le \mathrm{\pi }$

${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\mathrm{\pi }$ only if ${\sin }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{2}$ and ${\sin }^{-1}\left(y\right)=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \Rightarrow \mathrm{x}=\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \mathrm{y}=\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow {\left(\mathrm{x}+\mathrm{y}\right)}^2={\left(1+1\right)}^2=4 \end{gathered}$$

Hence value of ${\left(\mathrm{x}+\mathrm{y}\right)}^2$ is $4$.