Question

$se{c}^{-1}\frac{x}{a}-se{c}^{-1}\frac{x}{b}=se{c}^{-1}b-se{c}^{-1}a$, then 'x' is equal to

• A. a
• B. b
• C. ab
• D. 1

Answer 1

Answer: (C)

ab

$co{s}^{-1}\frac{a}{x}+co{s}^{-1}\frac{1}{a}=co{s}^{-1}\frac{a}{b}+co{s}^{-1}\frac{b}{x}$
$\therefore \frac{a}{x}.\frac{1}{a}-\sqrt{(1-\frac{a^2}{x^2})}\sqrt{(1-\frac{1}{a^2})}$
$=\frac{1}{b}.\frac{b}{x}=\sqrt{(1-\frac{b^2}{x^2})}\sqrt{(1-\frac{1}{b^2})}$
or $b^2(x^2-a^2)(a^2-1)=a^2(x^2-b^2)(b^2-1)$
or $x^2(a^2-b^2)=a^2{b}^2(a^2-b^2)\therefore x=ab$