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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Find $\cos\tan^{-1}\sin\cot^{-1}x$

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

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1 Answer

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$\cos\tan^{-1}\sin\cot x=\cos[\tan^{-1}(\sin(\sin^{-1}\large\frac{1}{\sqrt{1+x^2}}))]$ $if\;x\;>\;0$
$\qquad\qquad\qquad\quad=\cos[\tan^{-1}(\sin(\pi-\sin^{-1}\large\frac{1}{\sqrt{1+x^2}}))]$ $if\;x\;<\;0$
In each case
$\Rightarrow \cos\big[\tan^{-1}\large\frac{1}{\sqrt{1+x^2}}\big]$
$\Rightarrow \cos\big[\cos^{-1}\sqrt{\large\frac{1+x^2}{2+x^2}}\big]$
$\Rightarrow \sqrt{\large\frac{x^2+1}{x^2+2}}$
Hence (a) is the correct answer.
answered Oct 16, 2013 by sreemathi.v
 

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