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  4. If sec-1((x/a))- sec-1((x/b)) = sec-1b- sec-1a, then x =

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

Inverse Trigonometric Functions

Solution:

$sec^{-1} \left(\frac{x}{a}\right) -sec^{-1} \left(\frac{x}{b}\right) = sec^{-1}b - sex^{-1}a $
$ \Rightarrow 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}\cdot\frac{1}{a} - \sqrt{1-\frac{a^{2}}{x^{2}} } \sqrt{1-\frac{1}{a^{2}}}\right] $
$ = cos^{-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 $