December 31, 2024
Explanation:
(b)
logcosx sinx + logsinx cosx = 2
logcosx sinx + 1/logcosx sinx = 2
let logcosx sinx = k
k + 1/k = 2 [∵ k > 0]
k=1
logcosx sinx = k =1
sin x = (cos x)^1
sin x = cos x
Above relation is possible only in 1st quadrant.
∴ x = π/4 or 2π + π/4 or 4π +π/4 &…. so on
∴ x = 2nπ +π/4 Ans.