If $\sinh (2 \tanh ^{-1} x) = \frac{11}{60}$,then $x =$

Let $f:[0, 4\pi] \rightarrow [0, \pi]$ be defined by $f(x) = \cos^{-1}(\cos x)$. The number of points $x \in [0, 4\pi]$ satisfying the equation $f(x) = \frac{10-x}{10}$ is

If $\theta = 2 \tan^{-1} \frac{1}{8} + 2 \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}$ and $\tan \frac{\theta}{2} = \sqrt{m} + \sqrt{n}$,where $m$ and $n$ are positive integers such that $m < n$,then $(m^n + n^m)^{m+n}$ is equal to

If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$,then for $0 < a < \frac{1}{\sqrt{3}}$,the value of $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ is

If $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} ( \theta )$,then $\theta =$

Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \csc x)$