Trigonometric Identities Question 118
Question: $ \tan 1{}^\circ \tan 2{}^\circ \tan 3{}^\circ \tan 4{}^\circ ……..\tan 89{}^\circ = $
[MP PET 1998, 2001; AMU 1999; Pb. CET 1994]
Options:
A) 1
B) 0
C) $ \infty $
D) 1/2
Show Answer
Answer:
Correct Answer: A
Solution:
$ \tan 1{}^\circ \tan 2{}^\circ ….\tan 89{}^\circ $ $ =(\tan 1{}^\circ \tan 89{}^\circ )(\tan 2{}^\circ \tan 88{}^\circ )…… $ $ =1\times 1\times 1….=1. $
BETA
