Explanation

(a)
Let $A = \frac{1}{3\sin{\theta - 4\cos{\theta + 7}}}$
Now, $A$ will be minimum when $3\sin{\theta - 4\cos{\theta + 7}}$ is maximum
$\therefore$ Maximum value of
$3\sin{\theta - 4\cos{\theta + 7 = \sqrt{3^{2} + 4^{2}} + 7} = 12}$
$\therefore$ Minimum value of $\frac{1}{3\sin{\theta - 4\cos{\theta + 7}}}$ is $\frac{1}{12}$