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