A) \[\frac{5}{12}\]
B) \[\frac{5}{4}\]
C) \[\frac{60}{169}\]
D) \[\frac{169}{60}\]
Correct Answer: B
Solution :
| \[\frac{\tan \theta }{1+{{\tan }^{2}}\theta }=\frac{\tan \theta }{{{\sec }^{2}}\theta }\] |
| \[=\frac{\sin \theta }{\cos \theta }\times {{\cos }^{2}}\theta\] \[\left[ \because \,\tan \theta =\frac{\sin \theta }{\cos \theta } \right]\] |
| \[\left[ \because \,1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \right]\] |
| \[=\sin \theta \times \cos \theta\] |
| \[=\frac{PR}{PQ}\times \frac{RQ}{PQ}=\frac{9}{15}\times \frac{12}{15}\] |
| \[\left[ PR=\sqrt{225-144}=\sqrt{81}=9 \right]\] |
| \[=\frac{12}{25}\] |
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