iitkgp_bipin  is right but more detailed solution is given below so that you can easily solve any question of this type.

Let                         cosec - sin= m  so 1/sin- sin= m  

                               Þ(1-Sin2) /sin= m

                               Þ Cos2/ sin= m .............. (1)

Sameway as above the sec      - cos      = n   becomes

                               Sin2/Cos= n     ............... (2)

Multiplying (1) and (2)      ( Cos2/ sin) * (Sin2/Cos) = mn

                                              Þ SinCos= mn  ...................(3)

Now again from (1)             Cos2 /sin= m

                                              Þ Cos2= m Sin

                                              Þ Cos3= m SinCos

                                              Þ Cos3= m2n                                  [Using (3)]

                                              Þ Cos= m2/3 n1/3     ............(4)          

Same way from (2) and (3) you can get

                                              Þ Sin3= mn2      

                                               Þ Sin= m1/3 n2/3     .................(5)

Squaring (4) and (5) and adding you will get

               Cos2+ Sin2= (m2/3 n1/3 )2 + (m1/3 n2/3 )2   

               Þ (m2/3 n1/3 )2 + (m1/3 n2/3 )2   = 1

                Þ (m2n)2/3 + (mn2)2/3 = 1

So answer is 1