IIT JEE 2011 QUESTION

multiply the numerator and the denominator by 2

numerator will become 2               &

the denominator             =   2sin^2a+6sinacosa+10cos^2a

    taking a (sin^2a + cos^2a) aside  we get

                                  =   sin^2a +6sinacosa +9cos^2a +sin^2a + cos^2a

                                  =   (sina + 3 cosa)^2  + sin^2a +cos^2a

                                  =  (sina + 3 cosa)^2  + 1

now the expression becomes                      2/( (sina + 3 cosa)^2  + 1  )

for the maximum value of the expression  the denominator must be minimum 

and for denominator to be minimum , (sina + 3 cosa)^2  must be minimum 

the minimum value of , (sina + 3 cosa)^2  can be zero   when   tana = -3

therefore the maximum value of the expression        2/( (sina + 3 cosa)^2  + 1  ) =  2/(  0 +1 )

                                                                    = 2                   

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Dear Smit,

Sanatan is right!! Its the right answer.

cosec^2(a)+5(cos^2(a))