If secA=5/4, then the value of tanA/(1+tan^2A)-sinA/secA is:

secA = \(\frac{5}{4}\) = \(\frac{H}{B}\)

Using triplet ( 3 , 4 , 5) , We find that P = 3

Now , $\frac{tanA}{1+tan^2A}-\frac{sinA}{secA}$

= \(\frac{3/4}{1 + 9/16}\) - \(\frac{3 × 4 }{ 5 × 5}\) 

= \(\frac{12}{25}\) - \(\frac{12 }{ 25}\) 

= 0