If k sin²x + (1/k) cosec ² x = 2, for all x belonging to (0, pie/2) then cos²x + 5 sinx cosx + 6 sin ²x is equals to

(a) (k²+5k+6) / k²

(b) (k²-5k+6) / k²

(c)6

(d) none of these

as k sin²x + (1/k) cosec ² x = 2

which is (ksin²x - 1)² = 0

or sin²x = 1/k  , cos²x = 1 -(1/k)

for cos²x + 5 sinx cosx + 6 sin ²x

= 5 - 5/k +6/k + 5/k*(k-1)^1/2

= ( 1+5k+ 5*(k-1)^1/2) / k

ans: none of the these

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IIT Kgp - 05 batch