$x, y, z$ are in $G$.$P$. and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,then

$\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))$ is equal to :
(The inverse trigonometric functions take the principal values)

If $y = \sum_{k=1}^{6} k \cos^{-1} \left\{ \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right\}$,then $\frac{dy}{dx}$ at $x = 0$ is

$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots$ up to $\infty$ is equal to

Let $x = \frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos(2 \sin^{-1} x) = \frac{1}{9}$ and let $\alpha, \beta$ $(\alpha > \beta)$ be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line

For $k \in R$,let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k$,where $0 < |x| < \frac{1}{\sqrt{2}}$,be $\alpha$ and $\beta$,where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2}- bx -5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$,then $\frac{b}{k^{2}}$ is equal to $......$