In triangle ABC, we are given the expression cosec A (sin B cos C + cos B sin C) and we need to simplify it.

First, let's simplify the expression inside the brackets:

sin B cos C + cos B sin C = sin(B+C)

Using the trigonometric identity sin(A+B) = sin A cos B + cos A sin B, we can rewrite sin(B+C) as sin B cos C + cos B sin C.

Therefore, the expression inside the brackets simplifies to sin(B+C).

Now, let's substitute this simplified expression back into the original expression:

cosec A (sin B cos C + cos B sin C) = cosec A sin(B+C)

Next, let's use the trigonometric identity sin(A+B) = sin A cos B + cos A sin B to rewrite sin(B+C) as sin B cos C + cos B sin C:

cosec A (sin B cos C + cos B sin C) = cosec A sin B cos C + cosec A cos B sin C

Now, since cosec A is equal to 1/sin A, we can rewrite the expression as:

cosec A (sin B cos C + cos B sin C) = (1/sin A) sin B cos C + (1/sin A) cos B sin C

Using the distributive property, we can simplify further:

cosec A (sin B cos C + cos B sin C) = (sin B cos C)/sin A + (cos B sin C)/sin A

Finally, using the trigonometric identity cosec A = 1/sin A, we can rewrite the expression as:

cosec A (sin B cos C + cos B sin C) = (sin B cos C)/sin A + (cos B sin C)/sin A = sin B cos C cosec A + cos B sin C cosec A = 1 + 1 = 2

Therefore, the expression cosec A (sin B cos C + cos B sin C) simplifies to 2.

Since the options provided do not include 2 as a possible answer, the correct answer must be none of these (d).