A) \[\left( \frac{1+{{p}^{2}}}{2p} \right)\]
B) \[\left( \frac{2p}{1+{{p}^{2}}} \right)\]
C) \[\left( \frac{1+p}{2p} \right)\]
D) \[\left( \frac{1-p}{2p} \right)\]
E) \[\left( \frac{1-{{p}^{2}}}{2p} \right)\]
Correct Answer: E
Solution :
Given that, \[tan\text{ }20{}^\circ =p\] \[\frac{\tan 160{}^\circ -\tan 110{}^\circ }{1+\tan 60{}^\circ \tan 110{}^\circ }\] \[=\frac{\tan (180{}^\circ -20{}^\circ )-\tan (90{}^\circ +20{}^\circ )}{1+\tan (180{}^\circ -20{}^\circ )\tan (90{}^\circ +20{}^\circ )}\] \[=\frac{-\tan 20{}^\circ +\cot 20{}^\circ }{1+\tan 20{}^\circ \cot 20{}^\circ }\] \[=\frac{-p+\frac{1}{p}}{1+1}\] \[=\frac{1-{{p}^{2}}}{2p}\]
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