The value of $\cos ^{-1} \left[ \cot \left( \sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} \right) + \cos ^{-1} \left( \frac{\sqrt{12}}{4} \right) + \sec ^{-1} \sqrt{2} \right]$ is

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$,then $8x^2 + 1 =$

The mean of numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.80$. Then $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} =$

Considering the principal values of inverse trigonometric functions,the value of the expression $\tan\left(2 \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to:

If $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the correct option$(s)$ is(are):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$

Let $E_1 = \{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$ and $E_2 = \{x \in E_1 : \sin^{-1}(\log_e(\frac{x}{x-1})) \text{ is a real number}\}$. (Here,the inverse trigonometric function $\sin^{-1} x$ assumes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Let $f : E_1 \rightarrow R$ be the function defined by $f(x) = \log_e(\frac{x}{x-1})$ and $g : E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Match the items in $LIST I$ with $LIST II$.

$LIST I$	$LIST II$
$P$. The range of $f$ is	$1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains	$2$. $(0, 1)$
$R$. The domain of $f$ contains	$3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is	$4$. $(-\infty, 0) \cup (0, \infty)$
	$5$. $(-\infty, \frac{e}{e-1}]$
	$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$