Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 $

Can you answer this question?

1 Answer

0 votes
$t= \tan \theta$
$\sqrt {1+t^2} =\sec \theta$
$x= \theta$
$y= \sin ^{-1} \cos \theta$
$\quad= \sin ^{-1} (\sin (\large\frac{\pi}{2} -$$\theta))$
$y= \large\frac{\pi}{2} -$$\theta$
$y= \large\frac{\pi}{2} $$-x$
or $\large\frac{dx}{dt} $$=-\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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App