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#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 2  Maths Textbook Solution.

$x= ab$
Given:
$\cos^{-1}\frac{a}{x}-\cos^{-1}\frac{b}{x}= \cos^{-1}\frac{1}{b}-\cos^{-1}\frac{1}{a}$
Hint:
First, we will separate the angle of $a$ terms then we will apply the formula,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
$\cos^{-1}\frac{a}{x}+\cos^{-1}\frac{1}{a} = \cos^{-1}\frac{1}{b}+\cos^{-1}\frac{b}{x}$
Solution:
We know that,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
Substituting the values in the formula, we get
$\cos^{-1}\left [ \frac{1}{x}-\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}\right ]= \cos^{-1}\left [ \frac{1}{x}\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}} \right ]$
$\Rightarrow \frac{1}{x}-\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}=\frac{1}{x}-\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}}$
$\Rightarrow\sqrt{1-\left ( \frac{a}{x} \right )^{2}} \sqrt{1-\left ( \frac{1}{a} \right )^{2}}=\sqrt{1-\left ( \frac{b}{x} \right )^{2}}\sqrt{1-\left ( \frac{1}{b} \right )^{2}}$
Squaring on both sides, we get
$\Rightarrow \left ( 1-\left ( \frac{a}{x} \right )^{2}\right )\left ( 1-\left ( \frac{1}{a} \right )^{2} \right )=\left ( 1-\left ( \frac{b}{x} \right ) ^{2}\right )\left ( 1-\left ( \frac{1}{b}\right )^{2} \right )$
$\Rightarrow 1-\left ( \frac{a}{x} \right )^{2}-\left ( \frac{1}{a} \right )^{2}+\left ( \frac{1}{x} \right )^{2}= 1-\left ( \frac{b}{x} \right )^{2}-\left ( \frac{1}{b} \right )^{2}+\left ( \frac{1}{x} \right )^{2}$
$\Rightarrow \left ( \frac{b}{x} \right )^{2}-\left ( \frac{a}{x} \right )^{2}= \left ( \frac{1}{a} \right )^{2}-\left ( \frac{1}{b} \right )^{2}$
On Simplifying, we get
$\Rightarrow \left ( b^{2} -a^{2}\right )a^{2}b^{2}= x^{2}\left ( b^{2}-a^{2} \right )$
$\Rightarrow x^{2}= a^{2}b^{2}$
$\Rightarrow x= ab$
Hence the result.

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