Trigonometric Identities Question 27
Question: $ \tan 100{}^\circ +\tan 125{}^\circ +\tan 100{}^\circ \tan 125{}^\circ = $
[DCE 1999]
Options:
A) 0
B) 1/2
C) -1
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
$ \tan ,(100^{o}+125^{o})=\frac{\tan ,100^{o}+\tan ,125^{o}}{1-\tan ,100^{o},\tan ,125^{o}} $
$ \therefore $ $ \tan ,225^{o}=\frac{\tan 100^{o}+\tan 125^{o}}{1-\tan ,100^{o},\tan ,125^{o}} $ i.e., $ 1=\frac{\tan ,100^{o}+\tan 125^{o}}{1-\tan 100^{o},\tan 125^{o}} $ i.e., $ \tan 100^{o}+\tan 125^{o}+\tan 100^{o}\tan 125^{o}=1. $
BETA
