Let $H(x)=3 x^4+6 x^3-2 x^2+1$ and $g(x)$ be a polynomial of degree one. If

$\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$ then

$H(-1)+2 H(2)-3 H(1)=$

The smallest positive value of $x$ (in degrees) for which $\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan (x) \tan \left(x-50^{\circ}\right)$ is

For $n \in \mathbf{N}$, if $f(n)=(\cos n x)(\sec x)^n$ and $g(n)=(\sin n x)(\sec x)^n$, then $f(2020)-f(2019)+(\tan x) g(2019)=$

$\theta$ and $\alpha$ lie in $Q_3$. If $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression, then $\cos \theta \sec \frac{\alpha}{2}=$