If $secA=\frac{5}{4}$, then the value of $\frac{tanA}{1+tan^2A}-\frac{sinA}{secA}$ is:

2 1 0 3

0

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