Using standard trigonometric identities for powers of cosecant and cotangent, what is the simplified numerical value of the expression cosec^6 A − cot^6 A − 3 cosec^2 A cot^2 A?

Introduction / Context:This question focuses on algebraic manipulation of higher powers of trigonometric functions, specifically cosec A and cot A. By recognising a pattern of cubes and using the identity relating cosec^2 A and cot^2 A, you can simplify the expression to a simple constant. This type of problem checks both algebraic insight and familiarity with trigonometric identities.

Given Data / Assumptions:

• The expression is E = cosec^6 A − cot^6 A − 3 cosec^2 A cot^2 A.
• Angle A is such that cosec A and cot A are defined and finite.
• We must simplify E to a single numerical value if possible.

Concept / Approach:Notice that cosec^6 A and cot^6 A can be viewed as cubes of cosec^2 A and cot^2 A. This suggests using the algebraic identity a^3 − b^3 = (a − b)(a^2 + ab + b^2). Also, there is an important trigonometric identity: cosec^2 A − cot^2 A = 1. Introducing variables u = cosec^2 A and v = cot^2 A allows us to rewrite the expression in a purely algebraic form, which simplifies nicely.

Step-by-Step Solution:Let u = cosec^2 A and v = cot^2 A. Then cosec^6 A = u^3 and cot^6 A = v^3, and cosec^2 A cot^2 A = uv.The expression becomes E = u^3 − v^3 − 3uv.Use the identity for difference of cubes: u^3 − v^3 = (u − v)(u^2 + uv + v^2).So E = (u − v)(u^2 + uv + v^2) − 3uv.We know from trigonometric identities that cosec^2 A − cot^2 A = 1, so u − v = 1.Thus E = (u^2 + uv + v^2) − 3uv = u^2 − 2uv + v^2 = (u − v)^2.Since u − v = 1, we have E = (1)^2 = 1.

Verification / Alternative check:Choose a particular angle A where sine and cosine are simple, for example A = 45 degrees. Then sin 45° = sqrt(2)/2, so cosec 45° = sqrt(2), and cot 45° = 1. Compute E numerically: cosec^2 45° = 2, cot^2 45° = 1. Then cosec^6 45° = 8 and cot^6 45° = 1. Also, 3 cosec^2 45° cot^2 45° = 3 * 2 * 1 = 6. So E = 8 − 1 − 6 = 1. This matches the algebraic result and confirms the simplification.

Why Other Options Are Wrong:

• Values −2, −1, 0, or 2 do not appear from the algebraic transformation using the correct identities.
• These incorrect values usually arise from mishandling the difference of cubes identity or misusing the identity cosec^2 A − cot^2 A = 1.
• Only 1 corresponds to (u − v)^2 when u − v = 1.

Common Pitfalls:

• Failing to see that cosec^6 A and cot^6 A can be expressed as u^3 and v^3 with u = cosec^2 A and v = cot^2 A.
• Using the wrong identity, for example confusing cosec^2 A − cot^2 A with some other relation.
• Forgetting to subtract 3uv after expanding u^3 − v^3, which prevents the expression from collapsing into a perfect square.

Final Answer:1