Q. If A (0,/2) then (x²+x)^1/2 + tan²A       is always greater than or equal to

                                                                (x²+x)^1/2

      1) 2tanA

     2)1

     3)2

     4)sec²A

(x^2+x)^1/2+(tan A)*(tan A)/(x^2+x)^1/2=

((x^2+x)+tanA*tanA)/(x^2+x)^1/2

By AM>GM

((x^2+x)+tan A*tanA)/2>((x^2+x)*tanA*tanA))^1/2

therefore,

((x^2+x)+tanA*tanA)/(x^2+x)^1/2>2*tanA

Answer=(a)2 tan A