Question

Solve and find the value of $x$.

${\sin }^{-1}\left(\frac{2p}{1+p^2}\right)-{\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Answer 1

$\because {\sin }^{-1}\left(\frac{2p}{1+p^2}\right)=2{\tan }^{-1}p$ and ${\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)=2{\tan }^{-1}q$

$\therefore {\sin }^{-1}\left(\frac{2p}{1+p^2}\right)+{\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)=2{\tan }^{-1}p+2{\tan }^{-1}q=2{\tan }^{-1}\left(\frac{p+q}{1-pq}\right)$

Thus, $\frac{2x}{1-x^2}=\frac{p+q}{1-pq}\Rightarrow p+q=2x$ and $pq=x^2$

$\because (p-q{)}^2=(p+q{)}^2-4pq=(2x{)}^2-4x^2=4x^2-4x^2=0\Rightarrow (p-q{)}^2=0\Rightarrow p=q$

Thus, $p=q=x$