Question

The equation of the tangent to the curve
$y={\sin }^{-1}\frac{2x}{1+x^2}$ at $x=\sqrt{3}$

• A. $y=-\frac{x}{2}$
• B. $y-\frac{\pi }{3}=-\frac{x}{2}$
• C. $y-\frac{\pi }{3}=\frac{1}{2}(x-\sqrt{3})$
• D. $y-\frac{\pi }{3}=\frac{-1}{2}(x-\sqrt{3})$

Answer 1

The correct option is D $y-\frac{\pi }{3}=\frac{-1}{2}(x-\sqrt{3})$

$y={\sin }^{-1}\frac{2x}{1+x^2}$
Let $x=\tan \theta$
$\Rightarrow y={\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta }={\sin }^{-1}\left(\sin 2\theta \right)$
$={\sin }^{-1}\left(\sin 2\theta \right)=\pi -2\theta$
$(\because {\sin }^{-1}(\sin x)=\pi -x,x\in (\frac{\pi }{2},\frac{3\pi }{2}))$
$\therefore y=\pi -2{\tan }^{-1}x$
$\Rightarrow \frac{dy}{dx}=-\frac{2}{1+x^2}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\sqrt{3}}=-\frac{2}{1+3}=-\frac{1}{2}$
Also when $x=\sqrt{3}$, $y=\pi -2\cdot \frac{\pi }{3}=\frac{\pi }{3}$
Hence, equation of tangent is
$y-\frac{\pi }{3}=-\frac{1}{2}(x-\sqrt{3})$