A) \[\frac{9}{25}\]
B) \[\frac{12}{25}\]
C) \[\frac{3}{4}\]
D) \[\frac{1}{25}\]
Correct Answer: B
Solution :
\[\because \] \[\sec \alpha =\frac{5}{4}\]. \[\therefore \]\[\tan \alpha =\sqrt{{{\sec }^{2}}\alpha -1}\] \[\therefore \]\[=\sqrt{\frac{25}{16}-1}=\sqrt{\frac{25-16}{16}}=\sqrt{\frac{9}{16}}=\frac{3}{4}\] Now, \[\frac{\tan \alpha }{1+{{\tan }^{2}}\alpha }=\frac{3/4}{1+{{(3/4)}^{2}}}=\frac{3/4}{1+9/16}\] \[=\frac{3/4}{25/16}=\frac{12}{25}\]
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