Trigonometric Identities Question 97
Question: The value of $ \frac{\tan 70^{o}-\tan 20^{o}}{\tan 50^{o}}= $
[Karnataka CET 2003]
Options:
A) 1
B) 2
C) 3
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\tan 70^{o}-\tan 20^{o}}{\tan 50^{o}} $ = $ \frac{\frac{\sin 70^{o}}{\cos 70^{o}}-\frac{\sin 20^{o}}{\cos 20^{o}}}{\frac{\sin 50^{o}}{\cos 50^{o}}} $ = $ \frac{\frac{\sin 70^{o}\cos 20^{o}-\cos 70^{o}\sin 20^{o}}{\cos 70^{o}\cos 20^{o}}}{\frac{\sin 50^{o}}{\cos 50^{o}}} $ = $ \frac{2}{2}\times \frac{\sin (70^{o}-20^{o})\cos 50^{o}}{\cos 70^{o}\cos 20^{o}\sin 50^{o}} $ = $ \frac{2\sin 50^{o}\cos 50^{o}}{2\cos 70^{o}\cos 20^{o}\sin 50^{o}} $ = $ \frac{2\cos 50^{o}}{\cos 90^{o}+\cos 50^{o}}=\frac{2\cos 50^{o}}{0+\cos 50^{o}} $ = 2.
BETA
