What is the value of (3cos62^o)/(sin28^o)-(2tan34^o)/(cot56^o)?

The concept we use here is :-

sinθ = cos ( 90º - θ )

& tanθ = cot ( 90º - θ ) 

Now,

\(\frac{3 cos62º}{sin28º}\) - \(\frac{2 tan34º}{cot56º}\)

= \(\frac{3 sin28º}{sin28º}\) - \(\frac{2 tan34º}{tan34º}\)

= 3 -2

= 1