Trigonometric Identities Question 58
Question: If $ x=\cos 10{}^\circ \cos 20{}^\circ \cos 40{}^\circ , $ then the value of $ x $ is
[Roorkee 1995]
Options:
A) $ \frac{1}{4}\tan 10{}^\circ $
B) $ \frac{1}{8}\cot 10{}^\circ $
C) $ \frac{1}{8}cosec10{}^\circ $
D) $ \frac{1}{8}\sec 10{}^\circ $
Show Answer
Answer:
Correct Answer: B
Solution:
$ x=\cos 10^{o},\cos 20^{o}\cos 40^{o} $ $ =\frac{1}{2\sin 10^{o}},[2\sin 10^{o}\cos 10^{o}\cos 20^{o}\cos 40^{o}] $ $ =\frac{1}{2,.,2\sin 10^{o}},[2\sin 20^{o}\cos 20^{o}\cos 40^{o}] $ $ =\frac{1}{2,.,4\sin 10^{o}}[2\sin 40^{o}\cos 40^{o})=\frac{1}{8\sin 10^{o}}(\sin 80^{o}) $ $ =\frac{1}{8\sin 10^{o}}\cos 10^{o}=\frac{1}{8}\cot 10^{o} $ .
BETA
