The expression $\frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ}$ simplifies to

For any positive integer $n$,let $S_n: (0, \infty) \rightarrow R$ be defined by $S_n(x) = \sum_{k=1}^n \cot^{-1}\left(\frac{1+k(k+1)x^2}{x}\right)$,where for any $x \in R$,$\cot^{-1} x \in (0, \pi)$ and $\tan^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following statements is (are) $TRUE$?
$(A)$ $S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1+11x^2}{10x}\right)$,for all $x > 0$
$(B)$ $\lim_{n \rightarrow \infty} \cot(S_n(x)) = x$,for all $x > 0$
$(C)$ The equation $S_3(x) = \frac{\pi}{4}$ has a root in $(0, \infty)$
$(D)$ $\tan(S_n(x)) \leq \frac{1}{2}$,for all $n \geq 1$ and $x > 0$

If $\sin x \cosh y = \cos \theta$ and $\cos x \sinh y = \sin \theta$,then $\sin^2 x + \cosh^2 y =$

All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \leq 1$ also satisfy the equation

If $\theta = \frac{\pi}{6}$ and $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$,then $\sinh(x) =$

If $\cos (\theta - \alpha ), \cos \theta$ and $\cos (\theta + \alpha )$ are in $H.P.$,then $\cos \theta \sec \frac{\alpha }{2}$ is equal to