Quantitative Aptitude Ques 1202

Question: Find $ \frac{\tan 18{}^\circ }{\cot 72{}^\circ }-\frac{\cot 72{}^\circ }{\tan 18{}^\circ } $

Options:

A) $ 2 $

B) $ 1 $

C) $ 0 $

D) $ \sqrt{2} $

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ \frac{\tan 18{}^\circ }{\cot 72{}^\circ }-\frac{\cot 72{}^\circ }{\tan 18{}^\circ }=\frac{\tan (90{}^\circ -72{}^\circ )}{\cot 72{}^\circ }-\frac{\cot (90{}^\circ -18{}^\circ )}{\tan 18{}^\circ } $ $ =\frac{\cot 72{}^\circ }{\cot 72{}^\circ }-\frac{\tan 18{}^\circ }{\tan 18{}^\circ } $ $ [\because \tan (90{}^\circ -\theta )=\cot \theta ,cot(90{}^\circ -\theta )=tan\theta ] $ $ =1-1=0 $
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