Question

Solve the equation ${\cos }^{-1}\frac{a}{x}-{\cos }^{-1}\frac{b}{x}={\cos }^{-1}\frac{1}{b}-{\cos }^{-1}\frac{1}{a}$

Answer 1

$$\begin{gathered} {\cos }^{-1}\frac{a}{x}-{\cos }^{-1}\frac{b}{x}={\cos }^{-1}\frac{1}{b}-{\cos }^{-1}\frac{1}{a} \\ \Rightarrow {\cos }^{-1}\frac{a}{x}+{\cos }^{-1}\frac{1}{a}={\cos }^{-1}\frac{1}{b}+{\cos }^{-1}\frac{b}{x} \\ \Rightarrow {\cos }^{-1}\left[\frac{a}{x}\times \frac{1}{a}-\sqrt{1-{\left(\frac{a}{x}\right)}^2}\sqrt{1-{\left(\frac{1}{a}\right)}^2}\right]={\cos }^{-1}\left[\frac{b}{x}\times \frac{1}{b}-\sqrt{1-{\left(\frac{b}{x}\right)}^2}\sqrt{1-{\left(\frac{1}{b}\right)}^2}\right]\left[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)\right] \\ \Rightarrow {\cos }^{-1}\left[\frac{1}{x}-\sqrt{1-\frac{a^2}{x^2}}\times \sqrt{1-\frac{1}{a^2}}\right]={\cos }^{-1}\left[\frac{1}{x}-\sqrt{1-\frac{b^2}{x^2}}\times \sqrt{1-\frac{1}{b^2}}\right] \\ \Rightarrow \frac{1}{x}-\sqrt{1-\frac{a^2}{x^2}}\times \sqrt{1-\frac{1}{a^2}}=\frac{1}{x}-\sqrt{1-\frac{b^2}{x^2}}\times \sqrt{1-\frac{1}{b^2}} \\ \Rightarrow \left(1-\frac{a^2}{x^2}\right)\left(1-\frac{1}{a^2}\right)=\left(1-\frac{b^2}{x^2}\right)\left(1-\frac{1}{b^2}\right) \\ \Rightarrow 1-\frac{1}{a^2}-\frac{a^2}{x^2}+\frac{1}{x^2}=1-\frac{1}{b^2}-\frac{b^2}{x^2}+\frac{1}{x^2} \\ \Rightarrow \frac{a^2-b^2}{x^2}=\frac{1}{b^2}-\frac{1}{a^2} \\ \Rightarrow \frac{a^2-b^2}{x^2}=\frac{a^2-b^2}{a^2{b}^2} \\ \Rightarrow x^2=a^2{b}^2 \\ \Rightarrow x=ab \end{gathered}$$