Trigonometric Identities Question 158
Question: The value of $ {e^{{\log_{10}}\tan 1{}^\circ +{\log_{10}}\tan 2{}^\circ +{\log_{10}}\tan 3{}^\circ +………..+{\log_{10}}\tan 89{}^\circ }} $ is
Options:
A) 0
B) e
C) 1/e
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ {e^{{\log_{10}}\tan 1^{o}+{\log_{10}}\tan 2^{o}+{\log_{10}},\tan 3^{o}+……….+{\log_{10}},\tan 89^{o}}} $ $ ={e^{{\log_{10}},(\tan 1^{o},\tan 2^{o}\tan 3^{o}…..\tan 89^{o})}}={e^{{\log_{10}}1}}=e^{o}=1 $
BETA
