To simplify trigonometric functions for angles, it's important to reduce the angles to their standard form, which lies between $0^\circ$ and $360^\circ$. Here's how we can simplify each expression:

(A) Simplify $\sin(2460^\circ)$

1. Reduce $2460^\circ$ to an equivalent angle between $0^\circ$ and $360^\circ$:

   We can reduce any angle by subtracting multiples of $360^\circ$ (since one full rotation is $360^\circ$):

   $2460^\circ \div 360^\circ = 6.8333$

   So, subtract $360^\circ \times 6 = 2160^\circ$ from $2460^\circ$:

   $2460^\circ - 2160^\circ = 300^\circ$

   Therefore:

   $\sin(2460^\circ) = \sin(300^\circ)$

2. Evaluate $\sin(300^\circ)$:

   $300^\circ$ lies in the fourth quadrant, where the sine function is negative. The reference angle for $300^\circ$ is $360^\circ - 300^\circ = 60^\circ$.

   Therefore:

   $\sin(300^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$

   So:

   $\sin(2460^\circ) = -\frac{\sqrt{3}}{2}$

(B) Simplify $\tan(-150^\circ)$

1. Reduce $-150^\circ$ to a positive angle:

   To convert a negative angle into a positive one, add $360^\circ$:

   $-150^\circ + 360^\circ = 210^\circ$

   Therefore:

   $\tan(-150^\circ) = \tan(210^\circ)$

2. Evaluate $\tan(210^\circ)$:

   $210^\circ$ lies in the third quadrant, where the tangent function is positive. The reference angle for $210^\circ$ is $210^\circ - 180^\circ = 30^\circ$.

   Therefore:

   $\tan(210^\circ) = \tan(30^\circ) = \frac{1}{\sqrt{3}}$

   So:

   $\tan(-150^\circ) = \frac{1}{\sqrt{3}}$

Would you like further details or have any questions?

Here are 5 related questions:

1. Simplify $\cos(750^\circ)$.
2. Simplify $\sin(-45^\circ)$.
3. Simplify $\tan(1020^\circ)$.
4. Find the exact value of $\cos(330^\circ)$.
5. Simplify $\cot(210^\circ)$.

Tip: For trigonometric functions, always reduce the angle to within $0^\circ$ to $360^\circ$ first to simplify calculations.