Question
If 2cosA + secA = 2√2 , 0° < < 90°, then the value
of 2(sec 4 A + cosec 4 A) is:Solution
2cosA + secA = 2√2
Or, 2cosA + (1/cosA) = 2√2
Or, 2cos 2 A + 1 = 2√2 × cosA
Or, 2cos 2 A – 2√2 × cosA + 1 = 0
Or, 2cos 2 A - √2cosA - √2cosA + 1 = 0
Or, √2 × cosA × (√2cosA – 1) – 1 × (√2cosA – 1) = 0
Or, (√2 × cosA – 1) 2  = 0
Or, √2 × cosA – 1 = 0
So, cosA = (1/√2) = cos45 o
So, A = 45 o
2(sec 4 A + cosec 4 A) = 2 × (sec 4 45 o  + cosec 4 45 o )
= 2 × [(√2) 4  + (√2) 4 ] = 2 × (4 + 4) = 2 × 8 = 16
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