A) \[\sqrt{3}\]
B) \[-\sqrt{3}\]
C) \[\sqrt{2}-1\]
D) \[1-\sqrt{2}\]
Correct Answer: B
Solution :
\[L=\frac{1-\tan 2{}^\circ \cot 62{}^\circ }{\tan 152{}^\circ -\cot 88{}^\circ }=\frac{1-\tan 2{}^\circ \cot (90-28){}^\circ }{\tan (180-28){}^\circ -\cot (90-2){}^\circ }\]\[\Rightarrow \,\,L=\frac{1-\tan 2{}^\circ \tan 28{}^\circ }{-\tan 28{}^\circ -\tan 2{}^\circ }=-\left[ \frac{1-\tan 2{}^\circ \tan 28{}^\circ }{\tan 2{}^\circ +\tan 28{}^\circ } \right]\] \[\Rightarrow \,\,L=-\frac{1}{\tan \,(2+28){}^\circ }=-\frac{1}{\tan 30{}^\circ }=-\sqrt{3}\] \[\left[ \because \,\,\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \right]\]
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