Question : If $\sec A=\frac{5}{4}$, then the value of $\frac{\tan A}{1+\tan ^2 A}-\frac{\sin A}{\sec A}$ is:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: 3
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Correct Answer: 0
Solution :
$\sec A=\frac{5}{4}$
$\cos A = \frac{1}{\sec A} = \frac{4}{5}$
$\sin A = \sqrt{1-\cos^2A}=\sqrt{1-(\frac{4}{5})^2}=\frac{3}{5}$
$\tan A = \frac{\sin A}{\cos A}=\frac{3}{4}$
$\therefore$ $\frac{\tan A}{1+\tan ^2 A}-\frac{\sin A}{\sec A}$
$= \frac{\frac{3}{4}}{1+(\frac{3}{4})^2}-\frac{\frac{3}{5}}{\frac{5}{4}}$
$= \frac{3\times 4}{16+9}-\frac{3\times4}{5\times 5}$
$=\frac{12}{25}-\frac{12}{25}$
$=0$
Hence, the correct answer is 0.
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