For any y in R,let cot ^{-1}(y) in(0, pi) and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $	an ^{-1}\left(\frac{6 y}{9-y^2}ight)+\cot ^{-1}\left(\frac{9-y^2}{6 y}ight)=\frac{2 \pi}{3}$.
Let $x = \frac{6y}{9-y^2}$.

Vedclass · Accepted Answer

$4 \sqrt{3}-6$

Vedclass · Answer

(C) Given equation: $ an ^{-1}\left(\frac{6 y}{9-y^2} ight)+\cot ^{-1}\left(\frac{9-y^2}{6 y} ight)=\frac{2 \pi}{3}$. Let $x = \frac{6y}{9-y^2}$. Then the equation is $ an^{-1}(x) + \cot^{-1}(\frac{1}{x}) = \frac{2\pi}{3}$. Case $I$: If $x > 0$,then $\cot^{-1}(\frac{1}{x}) = an^{-1}(x)$. The equation becomes $2 an^{-1}(x) = \frac{2\pi}{3} \Rightarrow an^{-1}(x) = \frac{\pi}{3} \Rightarrow x = \sqrt{3}$. $\frac{6y}{9-y^2} = \sqrt{3} \Rightarrow 6y = 9\sqrt{3} - \sqrt{3}y^2 \Rightarrow \sqrt{3}y^2 + 6y - 9\sqrt{3} = 0 \Rightarrow y^2 + 2\sqrt{3}y - 9 = 0$. Solving for $y$: $y = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = -\sqrt{3} \pm \sqrt{12} = -\sqrt{3} \pm 2\sqrt{3}$. Since $x > 0$,we need $y \in (0, 3)$,so $y = \sqrt{3}$. Case $II$: If $x < 0$,then $\cot^{-1}(\frac{1}{x}) = an^{-1}(x) + \pi$. The equation becomes $2 an^{-1}(x) + \pi = \frac{2\pi}{3} \Rightarrow 2 an^{-1}(x) = -\frac{\pi}{3} \Rightarrow an^{-1}(x) = -\frac{\pi}{6} \Rightarrow x = -\frac{1}{\sqrt{3}}$. $\frac{6y}{9-y^2} = -\frac{1}{\sqrt{3}} \Rightarrow 6\sqrt{3}y = -9 + y^2 \Rightarrow y^2 - 6\sqrt{3}y - 9 = 0$. Solving for $y$: $y = \frac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = 3\sqrt{3} \pm \sqrt{36} = 3\sqrt{3} \pm 6$. Since $x < 0$,we need $y \in (-3, 0)$,so $y = 3\sqrt{3} - 6$. Sum of solutions: $\sqrt{3} + 3\sqrt{3} - 6 = 4\sqrt{3} - 6$.

For any $y \in R$,let $\cot ^{-1}(y) \in(0, \pi)$ and $\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation $\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0 < |y| < 3$,is equal to

Similar Questions

If $\cot \left(\cos^{-1} x\right) = \sec \left(\tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)$,where $b > a > 0$,then $x =$

The number of real solutions of $\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2}$ is

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

If $x = \sin \left( 2 \tan^{-1} 2 \right)$ and $y = \sin \left( \frac{1}{2} \tan^{-1} \frac{4}{3} \right)$,then -

Vedclass Test Series

Exam Paper Generator

Online Exam Module