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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 


(d) none of these

7 years ago


Answers : (1)


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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7 years ago

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