Question

$y={\tan }^{-1}\frac{u}{\sqrt{1-u^2}}$ & $x={\sec }^{-1}\frac{1}{2u^2-1}$, $u\in (0,\frac{1}{\sqrt{2}})$ $\cup \left(\frac{1}{\sqrt{2}},1\right)$, prove that $2\frac{dy}{dx}+1=0$.

Answer 1

$y={\tan }^{-1}\frac{u}{\sqrt{1-u^2}}$ & $x={\sec }^{-1}\frac{1}{2u^2-1}$
Put $u=\sin \theta$ in the above equations.
Then,
$y={\tan }^{-1}\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$ & $x={\sec }^{-1}\frac{1}{2{\sin }^2\theta -1}$
$or,y={\tan }^{-1}\frac{\sin \theta }{\cos \theta }$ & $x={\sec }^{-1}\frac{1}{-\left(\cos 2\theta \right)}$
$or,y={\tan }^{-1}\tan \theta$ & $x={\sec }^{-1}(-\sec 2\theta )$
$or,y=\theta$ & $x=\pi -2\theta$
Now, $\frac{dy}{d\theta }=1$ & $\frac{dx}{d\theta }=-2$
$\therefore \frac{dy}{dx}=-\frac{1}{2}$
$or,2\frac{dy}{dx}+1=0$.