For $x \in \mathbb{R}$,the range of $3 \cos (4x - 5) + 4$ lies in the interval:

$a, b, c$ are the sides of a scalene triangle $ABC$. If angles $\alpha, \beta, \gamma$ lie between $0$ and $\pi$ such that $\cos \alpha = \frac{a}{b+c}, \cos \beta = \frac{b}{c+a}$ and $\cos \gamma = \frac{c}{a+b}$,then $\tan^2 \frac{\alpha}{2} + \tan^2 \frac{\beta}{2} + \tan^2 \frac{\gamma}{2} =$

The minimum value of $8 \cos^2 x + 18 \sec^2 x$ for all $x \in R$ wherever it is defined,is:

If $A+B+C=\frac{3 \pi}{2}$,then $4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C=$

If $p = \sin^2 \theta + \cos^4 \theta$,then for all real values of $\theta$,which of the following is true?

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to