A) \[\frac{1}{5}\]
B) \[\frac{9}{5}\]
C) \[\frac{12}{25}\]
D) \[\frac{25}{12}\]
Correct Answer: C
Solution :
Given, \[\tan \theta =\frac{3}{4}\] \[\therefore \] \[\sec \theta =\sqrt{1+{{\tan }^{2}}\theta }\] \[=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\] or \[\cot \theta =\frac{4}{5}\] \[\therefore \] \[\sin \theta \,\cos \theta =\tan \theta .{{\cos }^{2}}\theta \] \[=\frac{3}{4}\times \frac{16}{25}=\frac{12}{25}\]
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