# Solve the equation

Question:

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

Solution:

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

$\Rightarrow \cos ^{-1} \frac{a}{x}+\cos ^{-1} \frac{1}{a}=\cos ^{-1} \frac{1}{b}+\cos ^{-1} \frac{b}{x}$

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

$\Rightarrow \cos ^{-1}\left[\frac{1}{x}-\sqrt{1-\frac{a^{2}}{x^{2}}} \times \sqrt{1-\frac{1}{a^{2}}}\right]=\cos ^{-1}\left[\frac{1}{x}-\sqrt{1-\frac{b^{2}}{x^{2}}} \times \sqrt{1-\frac{1}{b^{2}}}\right]$

$\Rightarrow \frac{1}{x}-\sqrt{1-\frac{a^{2}}{x^{2}}} \times \sqrt{1-\frac{1}{a^{2}}}=\frac{1}{x}-\sqrt{1-\frac{b^{2}}{x^{2}}} \times \sqrt{1-\frac{1}{b^{2}}}$

$\Rightarrow\left(1-\frac{a^{2}}{x^{2}}\right)\left(1-\frac{1}{a^{2}}\right)=\left(1-\frac{b^{2}}{x^{2}}\right)\left(1-\frac{1}{b^{2}}\right)$

$\Rightarrow 1-\frac{1}{a^{2}}-\frac{a^{2}}{x^{2}}+\frac{1}{x^{2}}=1-\frac{1}{b^{2}}-\frac{b^{2}}{x^{2}}+\frac{1}{x^{2}}$

$\Rightarrow \frac{a^{2}-b^{2}}{x^{2}}=\frac{1}{b^{2}}-\frac{1}{a^{2}}$

$\Rightarrow \frac{a^{2}-b^{2}}{x^{2}}=\frac{a^{2}-b^{2}}{a^{2} b^{2}}$

$\Rightarrow x^{2}=a^{2} b^{2}$

$\Rightarrow x=a b$