Trigonometric Identities Question 219
Question: If $ x=\sin 130^{o},\cos 80^{o},y=\sin ,80^{o},\cos ,130^{o},z=1+xy, $ which one of the following is true
[AMU 1999]
Options:
A) $ x>0,y>0,z>0 $
B) $ x>0,y<0,0<z<1 $
C) $ x>0,y<0,z>1 $
D) $ x<0,y<0,,0<z<1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ x=\sin 130^{o}\cos 80^{o}, $ $ y=\sin 80^{o}\cos 130^{o} $
Þ $ x=\cos 40^{o}\cos 80^{o},,y=-\sin 80^{o}\sin 40^{o} $ So, $ x>0 $ and $ y<0and $ $ xy<0 $ Now $ z=1+xy $
Þ $ 0<z<1 $ .
BETA
