If A is an acute angle and tanA + cotA = 2, find the value of 7tan⁸A − 6cot⁸A + 8sec²A.

Correct — C. The stem is an image. It reads: if A is an acute angle and tanA + cotA = 2, find the value of 7tan⁸A − 6cot⁸A + 8sec²A.

tanA + cotA = 2 is tanA + 1⁄tanA = 2
Multiply by tanA: tan²A − 2tanA + 1 = 0
That is (tanA − 1)² = 0, so tanA = 1 and A = 45°

tan⁸45° = 1 and cot⁸45° = 1
sec²45° = 1 + tan²45° = 2
7(1) − 6(1) + 8(2) = 7 − 6 + 16 = 17 → option (c)

Why the others are wrong

• (a)
  7 — 7 is the 7tan⁸A term alone. The other two terms still have to be added, and at A = 45° they contribute −6 and +16.
• (b)
  16 — 16 is the 8sec²A term alone (8 × 2). The remaining 7tan⁸A − 6cot⁸A adds another +1, lifting the total to 17.
• (d)
  6 — 6 is only the coefficient of cot⁸A, and that term enters with a minus sign. The other two terms give 7 and 16, so no reading of the expression lands on 6.

The whole item turns on one fact: for a positive quantity t, t + 1⁄t is never less than 2, and it equals 2 only when t = 1.

So tanA + cotA = 2 does not merely allow A = 45° — it forces it. Once A is fixed, the eighth powers stop mattering, because 1 raised to any power is 1.

What is left is arithmetic on three standard values: tan⁸45° = 1, cot⁸45° = 1 and sec²45° = 2. The high exponents are dressing on a 45° question.

Key facts

• For t > 0, t + 1⁄t ≥ 2, with equality only at t = 1.
• tan45° = cot45° = 1, and sec45° = √2, so sec²45° = 2.
• sec²A = 1 + tan²A is the Pythagorean identity used to convert tanA into sec²A.
• At A = 45° the expression evaluates as 7 − 6 + 16 = 17.

Common traps

• Guessing tanA = 2 from tanA + cotA = 2 instead of solving the quadratic.
• Taking sec²45° as 1 rather than 2.
• Reading the exponent 8 as a multiplier, giving 7 × 8 in place of 7tan⁸A.

No directly related past PYQ was found.