29. If cose + seco = 2, then the value of cos8o + sino is
(1) 2
(2)
(3) 4
(4) 8
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Solution :-
→ cosA + secA = 2
→ cosA + (1/cosA) = 2
→ (cos²A + 1)/cosA = 2
→ cos²A + 1 = 2cosA
→ cos²A - 2cosA + 1 = 0
→ cos²A - 2 * 1 * cosA + (1)² = 0
comparing with a² - 2ab + b² = (a - b)² we get,
→ (cosA - 1)² = 0
→ cosA - 1 = 0
→ cosA = 1
Therefore,
→ cosA = cos0°
→ A = 0°
→ cos^8A + sin^8A
→ cos^8(0°) + sin^8(0°)
→ (1)^8 + (0)^8
→ 1 + 0
→ 1 (Ans.)
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