If sec ^{-1} frac{x}{a}-sec ^{-1} frac{x}{b}= | vedclass

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(A) Given: $\sec ^{-1} \left(\frac{x}{a}\right)-\sec ^{-1} \left(\frac{x}{b}\right)=\sec ^{-1} b-\sec ^{-1} a$Using $\sec ^{-1} z = \cos ^{-1} \left(\

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$a b$

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(A) Given: $\sec ^{-1} \left(\frac{x}{a}\right)-\sec ^{-1} \left(\frac{x}{b}\right)=\sec ^{-1} b-\sec ^{-1} a$Using $\sec ^{-1} z = \cos ^{-1} \left(\frac{1}{z}\right)$,we get:$\cos ^{-1} \left(\frac{a}{x}\right)-\cos ^{-1} \left(\frac{b}{x}\right)=\cos ^{-1} \left(\frac{1}{b}\right)-\cos ^{-1} \left(\frac{1}{a}\right)$Using the formula $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} \left(uv + \sqrt{1-u^2}\sqrt{1-v^2}\right)$,we equate the arguments:$\frac{a b}{x^2} + \sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac{b^2}{x^2}} = \frac{1}{a b} + \sqrt{1-\frac{1}{b^2}}\sqrt{1-\frac{1}{a^2}}$$\frac{a b}{x^2} + \frac{\sqrt{(x^2-a^2)(x^2-b^2)}}{x^2} = \frac{1}{a b} + \frac{\sqrt{(b^2-1)(a^2-1)}}{a b}$Multiplying by $x^2 ab$:$a^2 b^2 + ab\sqrt{(x^2-a^2)(x^2-b^2)} = x^2 + x^2\sqrt{(a^2-1)(b^2-1)}$$ab\sqrt{(x^2-a^2)(x^2-b^2)} - x^2\sqrt{(a^2-1)(b^2-1)} = x^2 - a^2 b^2$For the equation to hold,we set $x^2 - a^2 b^2 = 0$,which implies $x^2 = a^2 b^2$,so $x = ab$.

If $\sec ^{-1} \frac{x}{a}-\sec ^{-1} \frac{x}{b}=\sec ^{-1} b-\sec ^{-1} a$,then $x$ is equal to

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