If asin³ X + bcos³ X = sinXcosX and asinX = bcosX, then find the value of a² + b², provided that x is neither 0° nor 90°.

Correct — B. Use the second equation to replace b cos X inside the first.

Split off one power: a sin³X + b cos³X = (a sin X) × sin²X + (b cos X) × cos²X
Replace b cos X by a sin X: = a sin X × sin²X + a sin X × cos²X
Factor out a sin X: = a sin X (sin²X + cos²X) = a sin X

So a sin X = sin X cos X
Divide by sin X, which the proviso keeps non-zero: a = cos X

Then b cos X = a sin X = cos X sin X
Divide by cos X: b = sin X

a² + b² = cos²X + sin²X = 1 → option (b)

Why the others are wrong

• (a)
  a² - b² — a² − b² is an expression, not a value. With a = cos X and b = sin X it equals cos²X − sin²X, which changes as X changes.
• (c)
  0 — 0 would need a = b = 0. The working gives a = cos X and b = sin X instead, and cos²X + sin²X is 1.
• (d)
  a² + b² — a² + b² repeats what is asked for. The question wants its value, and eliminating X turns it into the number 1.

When two equations share the unknowns a and b, eliminate instead of solving for the angle: use the simpler equation to rewrite a term in the harder one.

Here a sin X = b cos X lets b cos X be swapped for a sin X. The cubes then factor round sin²X + cos²X = 1, and a and b come out as cos X and sin X.

The proviso that X is neither 0° nor 90° is what lets you divide by sin X and by cos X.

Key facts

• sin²X + cos²X = 1 for every angle X.
• Given a sin X = b cos X, the expression a sin³X + b cos³X reduces to a sin X.
• The two conditions together give a = cos X and b = sin X.
• So a² + b² = cos²X + sin²X = 1.

Common traps

• Trying to find X itself, when only a and b in terms of X are needed
• Writing a sin³X as a sin X × sin X, which loses a power and stops the factoring
• Choosing the option that restates a² + b² because it matches the question's wording

No directly related past PYQ was found.