In trigonometry, the angles A, B, and C have the values A = 30°, B = 60°, and C = 135°. Evaluate the expression (sin A)^3 + (cos B)^3 + (tan C)^3 − 3·sin A·cos B·tan C exactly.

Introduction / Context:
This question is designed to highlight a standard algebraic identity involving the sum of cubes. By combining this identity with exact trigonometric values for special angles, the expression becomes much easier to evaluate. The problem is a good example of how algebra and trigonometry work together in aptitude style questions, and it rewards recognition of structure instead of raw computation.

Given Data / Assumptions:
- Angle A is 30 degrees, so sin A is a standard special value.
- Angle B is 60 degrees, so cos B is also a standard special value.
- Angle C is 135 degrees, so tan C is a standard value in the second quadrant.
- We must compute (sin A)^3 + (cos B)^3 + (tan C)^3 − 3·sin A·cos B·tan C exactly.
- All calculations use exact surd values rather than decimal approximations.

Concept / Approach:
Recall the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). If a + b + c equals zero, then the entire expression is zero. In this problem, we can let a = sin A, b = cos B, and c = tan C. We compute each of these exactly using known trigonometric values and then check whether their sum is zero. If it is, the original expression immediately becomes zero by the identity.

Step-by-Step Solution:
1) Compute sin 30° = 1/2. 2) Compute cos 60° = 1/2. 3) Compute tan 135°. Since tan 135° = tan(180° − 45°) and tangent is negative in the second quadrant, tan 135° = −tan 45° = −1. 4) Set a = sin A = 1/2, b = cos B = 1/2, and c = tan C = −1. 5) Find a + b + c = 1/2 + 1/2 − 1 = 1 − 1 = 0. 6) Use the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). 7) Because a + b + c = 0, the entire expression equals 0, regardless of the second factor.

Verification / Alternative check:
We can compute the expression directly to confirm. First, a^3 = (1/2)^3 = 1/8 and b^3 = 1/8. Also c^3 = (−1)^3 = −1. Thus a^3 + b^3 + c^3 = 1/8 + 1/8 − 1 = 1/4 − 1 = −3/4. Next, the product 3abc = 3·(1/2)·(1/2)·(−1) = 3·(1/4)·(−1) = −3/4. So a^3 + b^3 + c^3 − 3abc = −3/4 − (−3/4) = 0. This matches the identity based result exactly.

Why Other Options Are Wrong:
Options A (1), B (8), D (9), and E (−1) are typical distractors that might result from partial calculations or ignoring the negative sign of tan 135°. For instance, forgetting that tangent is negative in the second quadrant can flip the sign of c and change the entire expression. Only Option C matches the exact value obtained using both the identity and direct computation.

Common Pitfalls:
The main pitfalls are using approximate decimal values instead of exact surds and overlooking the sign of tangent in different quadrants. Many students also forget or misuse the sum of cubes identity, attempting long and error prone expansions instead. Recognising the structure a^3 + b^3 + c^3 − 3abc and checking whether a + b + c vanishes is a powerful shortcut that prevents mistakes and saves time.

Final Answer:
Using exact special angle values and the sum of cubes identity, the given trigonometric expression simplifies to 0.