if tanA + secA = 2 where a is acute then which option is correct. 
a) tan A < Sec A
b) Cosec A > CotA
c) Sec A > Cosec A
d) CotA > Tan A

Your teacher made a mistake in asking 
"which option is correct"? He or she should
have asked "Which option is INcorrect?"
because it turns out that c) is the only 
INcorrect one!!

Put tanA + secA = 2 with the trig identity
1 + tan²A = sec²A

and we have the system of equations

tanA + secA = 2
  1 + tan²A = sec²A

Solve by substitution:
Solve the first for secA

       secA = 2 - tanA

Substitute into the second equation

  1 + tan²A = (2 - tanA)²
  1 + tan²A = 4 - 4tanA + tan²A

The tan²A's cancel and we get

          1 = 4 - 4tanA
      4tanA = 3
      tan A = 3/4

So we can draw a right triangle ABC 
containing angle A with 3 as A's 
opposite side and 4 as A's adjacent
side:

           B  
         . |
     .     |3 
A._________|
     4      C

We can calculate the hypotenuse by the 
Pythagorean theorem.

c² = a² + b²
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = 5

So the triangle has hypotenuse c = AB = 5

           B  
         . |
    5.     |3 
A._________|
     4      C

Now we check the choices:             

a) tan A < Sec A

   3/4 < 5/4   yes that's true 


b) Cosec A > CotA

    5/3 > 4/3  yes that's true


c) Sec A > Cosec A

    5/4 > 5/3  no that's false

d) CotA > Tan A
 
    4/3 > 3/4 yes that's true.

They are all true but c)!!

Edwin