Question
If 2cosA + secA = 2β2 , 0Β° < < 90Β°, then the value
of 2(sec 4 A + cosec 4 A) is:Solution
2cosA + secA = 2β2
Or, 2cosA + (1/cosA) = 2β2
Or, 2cos 2 A + 1 = 2β2 Γ cosA
Or, 2cos 2 A β 2β2 Γ cosA + 1 = 0
Or, 2cos 2 A - β2cosA - β2cosA + 1 = 0
Or, β2 Γ cosA Γ (β2cosA β 1) β 1 Γ (β2cosA β 1) = 0
Or, (β2 Γ cosA β 1) 2 Β = 0
Or, β2 Γ cosA β 1 = 0
So, cosA = (1/β2) = cos45 o
So, A = 45 o
2(sec 4 A + cosec 4 A) = 2 Γ (sec 4 45 o Β + cosec 4 45 o )
= 2 Γ [(β2) 4 Β + (β2) 4 ] = 2 Γ (4 + 4) = 2 Γ 8 = 16
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