Question

Let $x,y\in R$ satisfying the equation ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}(xy)=\frac{11\pi }{12}$, then the value of $\frac{dy}{dx}$ at $x=1$ is

• A. $-\left(\frac{3+\sqrt{3}}{3}\right)$
• B. $\left(\frac{1}{3}+\frac{1}{2\sqrt{3}}\right)$
• C. $-\left(\frac{5+\sqrt{3}}{3}\right)$
• D. $-\left(\frac{1}{3}+\frac{1}{2\sqrt{3}}\right)$

Answer 1

Answer: (D)

Given, ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}(xy)=\frac{11\pi }{12}$ .......(1)
Put $x=1$, we get
${\cot }^{-1}1+{\cot }^{-1}y+{\cot }^{-1}y=\frac{11\pi }{12}\Rightarrow 2{\cot }^{-1}y=\left(\frac{11\pi }{12}-\frac{\pi }{4}\right)=\frac{8\pi }{12}=\frac{2\pi }{3}\text{. }$
$\therefore {\cot }^{-1}y=\frac{\pi }{3}\Rightarrow y=\frac{1}{\sqrt{3}}$
So, $P\left(x=1,y=\frac{1}{\sqrt{3}}\right)$.
Now, differentiating both sides of equation (1) with respect to $x$, we get
$\frac{-1}{1+x^2}-\frac{1}{1+y^2}\cdot \frac{dy}{dx}-\frac{1}{1+x^2{y}^2}\cdot \left(x\cdot \frac{dy}{dx}+y\right)=0$
Put $x=1,y=\frac{1}{\sqrt{3}}$, we get
$\frac{-1}{2}-\left(\frac{1}{1+\frac{1}{3}}\right)\frac{dy}{dx}-\left(\frac{1}{1+\frac{1}{3}}\right)\left(1\cdot \frac{dy}{dx}+\frac{1}{\sqrt{3}}\right)=0$
$\Rightarrow \frac{-1}{2}-\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)=2\times \frac{3}{4}\cdot \left(\frac{dy}{dx}\right)$
$\Rightarrow -\left(\frac{2\sqrt{3}+3}{4\sqrt{3}}\right)=\frac{3}{2}\left(\frac{dy}{dx}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{-2}{3}\left(\frac{2\sqrt{3}+3}{4\sqrt{3}}\right)=\frac{-1}{3}-\frac{1}{2\sqrt{3}}=-\left(\frac{1}{3}+\frac{1}{2\sqrt{3}}\right)$