Considering only the principal values of the inverse trigonometric functions, the value of

$$\cot^{-1}(\cot(-11)) + 10 \sin\left(2 \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) + 10\sin(2 \tan^{-1}(2))$$

is

The total number of real solutions of the equation

$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1}\left(\frac{6 \tan \theta}{9 + \tan^2 \theta}\right) $

is

(Here, the inverse trigonometric functions $\sin^{-1} x$ and $\tan^{-1} x$ assume values in $[ -\frac{\pi}{2}, \frac{\pi}{2}]$ and $( -\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)

Considering only the principal values of the inverse trigonometric functions, the value of

$$ \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) $$

is

For any $y \in \mathbb{R}$, 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 :