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If $secA=\frac{5}{4}$, then the value of $\frac{tanA}{1+tan^2A}-\frac{sinA}{secA}$ is: |
2 1 0 3 |
0 |
secA = \(\frac{5}{4}\) = \(\frac{H}{B}\) 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 |
