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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

If

\sec^{-1} \frac{x}{a} - \sec^{-1} \frac{x}{b} = \sec^{-1} b \sec^{-1} a

x=

$-a b$
$a^{2}$
$b^{2}$
$b^{2}$

Solution:

\sec^{-1}\left(\frac{x}{a}\right) - \sec^{-1}\left(\frac{x}{b}\right)

= \sec^{-1} b - \sec^{-1} a

\Rightarrow \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)

\Rightarrow \cos^{-1}\left\{ \frac{a b}{x^{2}} + \sqrt{1-\frac{a^{2}}{x^{2}}} \cdot \sqrt{1-\frac{b^{2}}{x^{2}}} \right\}

= \cos^{-1}\left\{ \frac{1}{a b} + \sqrt{1-\frac{1}{b^{2}}} \sqrt{1-\frac{1}{a^{2}}} \right\}

[\because \cos^{-1} x - \cos^{-1} y = \cos^{-1}\left\{ x y + \sqrt{1-x^{2}} \sqrt{1-y^{2}} \right\}]

\Rightarrow \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}

\Rightarrow a^{2} b^{2} + a b \sqrt{(x^{2}-a^{2})(x^{2}-b^{2})}

= x^{2} + x^{2} \sqrt{(a^{2}-1)(b^{2}-1)}

\Rightarrow a b \sqrt{(x^{2}-a^{2})(x^{2}-b^{2})} - x^{2} \sqrt{(a^{2}-1)(b^{2}-1)} = x^{2} - a^{2} b^{2}

x^{2} - a^{2} b^{2}

in RHS

x^{2}

has the minimum value zero because

a^{2} b^{2}

is always positive and

x^{2} - a^{2} b^{2} = 0

is positive.

If

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

\Rightarrow x = a b

a b

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