The minimum value of $3\sin \theta + 4\cos \theta $ is

Let $M$ and $m$ respectively denote the maximum and the minimum values of $[f(\theta)]^2$,where $f(\theta)=\sqrt{a^2 \cos^2 \theta+b^2 \sin^2 \theta} + \sqrt{a^2 \sin^2 \theta+b^2 \cos^2 \theta}$. Then $M-m=$

Assertion $(A)$: If $\alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ}$,then $\tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma+\tan 2 \gamma \tan 2 \alpha=1$.
Reason $(R)$: In $\triangle ABC$,$\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$.
Which of the following is true?

If $A+B+C=\frac{\pi}{2}$,then $\sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1=$

The maximum value of $\sin(x + \pi/6) + \cos(x + \pi/6)$ is attained at $x =$

If $A+B+C=60^{\circ}$,then $\cos (30^{\circ}-A)+\cos (30^{\circ}-B)+\cos (30^{\circ}-C)+\sin (A+B+C) = $