Question

If ${\sec }^{-1}\left(\frac{x}{a}\right)-{\sec }^{-1}\left(\frac{x}{b}\right)$
$={\sec }^{-1}b-{\sec }^{-1}a$, then $x=$

• A. $ab$
• B. $\frac{a}{b}$
• C. $2ab$
• D. none of these

Answer 1

Answer: (A)

$se{c}^{-1}\left(\frac{x}{a}\right)-se{c}^{-1}\left(\frac{x}{b}\right)=se{c}^{-1}b-se{x}^{-1}a$
$\Rightarrow co{s}^{-1}\frac{a}{x}-co{s}^{-1}\frac{b}{x}=co{s}^{-1}\frac{1}{b}-co{s}^{-1}\frac{1}{a}$
$\Rightarrow co{s}^{-1}\frac{a}{x}+co{s}^{-1}\frac{1}{a}=co{s}^{-1}\frac{1}{b}-co{s}^{-1}\frac{b}{x}$
$\Rightarrow co{s}^{-1}\left[\frac{a}{x}\cdot \frac{1}{a}-\sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac{1}{a^2}}\right]$
$=co{s}^{-1}\left[\frac{1}{b}\cdot \frac{b}{x}-\sqrt{1-\frac{1}{b^2}}\sqrt{1-\frac{b^2}{x^2}}\right]$
$\Rightarrow \frac{1}{x}-\sqrt{1-\frac{a^2}{x^2}-\frac{1}{a^2}+\frac{1}{x^2}}$
$=\frac{1}{x}-\sqrt{1-\frac{1}{b^2}-\frac{b^2}{x^2}+\frac{1}{x^2}}$
$\Rightarrow \sqrt{1-\frac{a^2}{x^2}-\frac{1}{a^2}+\frac{1}{x^2}}$
$=\sqrt{1-\frac{b^2}{x^2}-\frac{1}{b^2}+\frac{1}{x^2}}$
$\Rightarrow 1-\frac{a^2}{x^2}-\frac{1}{a^2}+\frac{1}{x^2}$
$=1-\frac{b^2}{x^2}-\frac{1}{b^2}+\frac{1}{x^2}$
$\Rightarrow -\frac{a^2}{x^2}-\frac{1}{a^2}=-\frac{b^2}{x^2}-\frac{1}{b^2}$
$\Rightarrow \frac{b^2-a^2}{x^2}=\frac{1}{a^2}-\frac{1}{b^2}=\frac{b^2-a^2}{a^2{b}^2}$
$\Rightarrow x^2=a^2{b}^2$
$\Rightarrow x=ab$