Question: If $ \tan 15{}^\circ =2-\sqrt{3}, $ then the value of $ \tan 15{}^\circ \cot 75{}^\circ +\tan 75{}^\circ \cot 15{}^\circ $ is
Options:
A) 14
B) 12
C) 10
D) 8
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Answer:
Correct Answer: A
Solution:
- Given, $ \tan 15{}^\circ =2-\sqrt{3} $
Then, $ \tan 15{}^\circ \cdot \cot 75{}^\circ +tan75{}^\circ \cdot cot15{}^\circ $
$ =\tan 15{}^\circ \cdot \cot (90{}^\circ -15{}^\circ )+\tan (90{}^\circ -15{}^\circ )\cdot \cot 15{}^\circ $ $ ={{\tan }^{2}}15{}^\circ +{{\cot }^{2}}15{}^\circ $
(i)
$ [\because \tan (90{}^\circ -\theta )=\cot \theta ,\cot (90{}^\circ -\theta )=\tan \theta ] $
Now, $ \tan 15{}^\circ =2-\sqrt{3} $
$ \Rightarrow $ $ \cot 15{}^\circ =\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} $
$ =2+\sqrt{3} $ [on rationalisation]
$ \therefore $ $ {{\tan }^{2}}15{}^\circ +{{\cot }^{2}}15{}^\circ ={{(2-\sqrt{3})}^{2}}+{{(2+\sqrt{3})}^{2}} $
$ =[4+3-4\sqrt{3}]+[4+3+4\sqrt{3}] $
$ =7+7=14 $