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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\cot(\cos^{-1}x)=\sec\big[\tan^{-1}\large\frac{a}{\sqrt{b^2-a^2}}\big]$ then $x$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{b}{\sqrt{2b^2-a^2}}&(b)\;\large\frac{a}{\sqrt{2b^2-a^2}}\\(c)\;\large\frac{\sqrt{2b^2-a^2}}{a}&(d)\;\large\frac{\sqrt{2b^2-a^2}}{b}\end{array}$

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1 Answer

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Given :
$\cot(\cos^{-1}x)=\sec\big[\tan^{-1}\large\frac{a}{\sqrt{b^2-a^2}}\big]$
$\big[\cot^{-1}\big(\large\frac{x}{\sqrt{1-x^2}}\big)\big]=$$\sec\big[\sec^{-1}\large\frac{b}{\sqrt{b^2-a^2}}\big]$
$\Rightarrow \large\frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow x^2(b^2-a^2)=b^2-b^2x^2$
$x^2(2b^2-a^2)=b^2$
$x=\large\frac{b}{\sqrt{2b^2-a^2}}$
Hence (a) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
 

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