Trigonometric Identities Question 188
Question: What is $ \frac{1-\tan 2^{0}\cot 62^{0}}{\tan 152^{0}-\cot 88^{0}} $ equal to?
Options:
A) $ \sqrt{3} $
B) $ -\sqrt{3} $
C) $ \sqrt{2}-1 $
D) $ 1-\sqrt{2} $
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Answer:
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 }=-[ \frac{1-\tan 2{}^\circ \tan 28{}^\circ }{\tan 2{}^\circ +\tan 28{}^\circ } ] $
$ \Rightarrow L=-\frac{1}{\tan ,(2+28){}^\circ }=-\frac{1}{\tan 30{}^\circ }=-\sqrt{3} $ $ [ \because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} ] $
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