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If $x= \cos^{-1} \bigg( \large\frac{1}{\sqrt {1+t^2}}\bigg) $ and $\; y= \sin ^{-1} \bigg( \large\frac{1}{\sqrt {1+t^2}}\bigg) $ then $\large\frac{dy}{dx}$

$(a)\;\frac{-\sqrt {1+t^2}}{t} \\ (b)\;-1 \\ (c)\;\frac{\sqrt {1+t^2}}{t} \\ (d)\;1 $

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$t= \tan \theta$
$\sqrt {1+t^2} =\sec \theta$
$x= \theta$
$y= \sin ^{-1} \cos \theta$
$\quad= \sin ^{-1} (\sin (\large\frac{\pi}{2} -$$\theta))$
$y= \large\frac{\pi}{2} -$$\theta$
$y= \large\frac{\pi}{2} $$-x$
$\large\frac{dy}{dx}$$=-1$
or $\large\frac{dx}{dt} $$=-\large\frac{1}{\sqrt {1- \bigg( \Large\frac{1}{\sqrt {1+t^2}}\bigg)}}$
$\qquad=\large\frac{-1}{\sqrt {1- \Large\frac{1}{\sqrt {1+t^2}}}}$
$\large\frac{dx}{dt}=\large\frac{-1}{\sqrt {+2}}$$ \sqrt {1+t^2}$
$\large\frac{dy}{dt} $$=+ \large\frac{1}{\sqrt {1- \bigg(\Large\frac{1}{\sqrt {1+t^2}}\bigg)^2}}$
$\qquad= \large\frac{1 \sqrt {1+t^2}}{\sqrt {t^2}}$
$\large\frac{dy}{dx} =\frac{dy}{dt}. \frac{dt}{dx}$$=-1$
Hence b is the correct answer.
answered Jan 31, 2014 by meena.p
edited Mar 20, 2014 by balaji.thirumalai
 

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